We prove that all smooth Fano threefolds in the families
№2.1, №2.2, №2.3, №2.4, №2.6 and №2.7 are K-stable,
and we also prove that smooth Fano threefolds in the family №2.5 that satisfy one very explicit generality condition are K-stable.

§ INTRODUCTION

Let $X$ be a smooth Fano threefold. Then $X$ belongs to one of the $105$ families,
which are labeled as №1.1, №1.2, $\ldots$, №9.1, №10.1.
See [3], for the description of these families.
In $76$ families, K-polystable smooth Fano threefolds are classified [2, 3, 4, 7, 8, 12, 15, 17].
The remaining $29$ deformation families are

№1.9, №1.10, №2.1, $\ldots$, №2.7, №2.9, $\ldots$,
№2.21, №3.2, №3.4, $\ldots$, №3.8, №3.11.

General members of these $29$ families are K-polystable [3].
In this paper, we prove

Let $X$ be a smooth Fano threefold contained in one of the following deformation families:
№2.1, №2.2, №2.3, №2.4, №2.6, №2.7.
Then $X$ is K-stable.

Let $X$ be a smooth Fano threefold in the deformation family №2.5.
Recall that there exists the following Sarkisov link:

\xymatrix{

where $V$ is a smooth cubic threefold in $\mathbb{P}^4$,
the morphism $\pi$ is a blow up of a smooth plane cubic curve,
and $\phi$ is a morphism whose fibers are normal cubic surfaces.
Suppose that

\begin{equation}
\label{equation:generality-condition}\tag{$\bigstar$}
\text{no fiber of the morphism $\phi$ has a Du Val singular point of type $\mathbb{D}_5$ or $\mathbb{E}_6$.}
\end{equation}

Then $X$ is K-stable.

Let us describe the structure of this paper.
In Section <ref>, we prove Auxiliary Theorem,
and we prove that all smooth Fano threefolds in the families №2.1 and №2.3 are K-stable.
In Sections <ref>, <ref>, <ref>, <ref>,
we prove that all smooth Fano threefolds in the families №2.2,  №2.4,  №2.6, №2.7 are K-stable, respectively.
Note that Section <ref> is very technical and long.

In this paper, we use two applications of the Abban–Zhuang theory [1] which have been discovered in [3, 15].
For the background material, we refer the reader to [3, 15, 19].

Cheltsov has been supported by EPSRC grant №EP/V054597/1.
Fujita has been supported by JSPS KAKENHI Grant №22K03269.
This paper has been written during Cheltsov's two month stay at l'Institut des Hautes Études Scientifiques.
He thanks the institute for the hospitality and excellent working conditions.

§ FAMILIES №2.1, №2.3, №2.5

Fix $d\in\{1,2,3\}$. Let $V$ be one of the following smooth Fano threefolds:

$\boxed{d=1}$ a smooth sextic hypersurface in $\mathbb{P}(1,1,1,2,3)$;

d=2 a smooth quartic hypersurface in $\mathbb{P}(1,1,1,1,2)$;

d=3 a smooth cubic threefold in $\mathbb{P}^4$.

Then $-K_{V}\sim 2H$ for an ample divisor $H\in\mathrm{Pic}(V)$ such that $H^3=d$ and $\mathrm{Pic}(V)=\mathbb{Z}[H]$.
Let $S_1$ and $S_2$ be two distinct surfaces in the linear system $|H|$, and let $\mathcal{C}=S_1\cap S_2$.
Suppose that the curve $\mathcal{C}$ is smooth.
Then $\mathcal{C}$ is an elliptic curve by the adjunction formula.
Let $\pi\colon X\to V$ be the blow up of the curve $\mathcal{C}$, and let $E$ be the $\pi$-exceptional surface.

* If $d=1$, then $X$ is a smooth Fano threefold in the deformation family №2.1.

* If $d=2$, then $X$ is a smooth Fano threefold in the deformation family №2.3.

* If $d=3$, then $X$ is a smooth Fano threefold in the deformation family №2.5.

Moreover, all smooth Fano threefolds in these families can be obtained in this way.

Note that $(-K_X)^3=4d$. Moreover, we have the following commutative diagram:

\xymatrix{

where $V\dasharrow\mathbb{P}^1$ is the rational map given by the pencil that is generated by  $S_1$ and $S_2$,
and $\phi$ is a morphism whose general fiber is a smooth del Pezzo surface of degree $d$.

The goal of this section is to show that $X$ is K-stable in the case when $d=1$ or $d=2$,
and to show that $X$ is K-stable in the case when $d=3$ and $X$ satisfies the condition (<ref>).
To show that $X$ is K-stable,  it is enough to show that $\delta_O(X)>1$ for every point $O\in X$.
This follows from the valuative criterion for K-stability [14, 16].

Let $O$ be a point in $X$, let $A$ be the fiber of the morphism $\phi$ such that $O\in A$.
Suppose that $A$ has at most Du Val singularities at the point $O$. Then

\delta_O(X)\geqslant\left\{\aligned
&\mathrm{min}\Big\{\frac{16}{11},\frac{16}{15}\delta_O(A)\Big\}\ \text{if $O\not\in E$}, \\
&\mathrm{min}\Big\{\frac{16}{11},\frac{16\delta_{O}(A)}{\delta_{O}(A)+15}\Big\}\ \text{if $O\in E$}.
\endaligned
\right.

Let $u$ be a non-negative real number. Then $-K_X-uA\sim_{\mathbb{R}} (2-u)A+E$,
which implies that divisor $-K_X-uA$ is pseudoeffective if and only if $u\leqslant 2$.
For every $u\in[0,2]$, let us denote by $P(u)$ the positive part of Zariski decomposition of the divisor $-K_X-uA$,
and let us denote by $N(u)$ its negative part. Then

&(2-u)A+E\ \text{if $0\leqslant u\leqslant 1$}, \\
&(2-u)H\ \text{if $1\leqslant u\leqslant 2$},
\endaligned
\right.

&0\ \text{if $0\leqslant u\leqslant 1$}, \\
&(u-1)E\ \text{if $1\leqslant u\leqslant 2$}.
\endaligned
\right.

Integrating, we get $S_X(A)=\frac{11}{16}$.
Using <cit.> and <cit.>, we get

\begin{equation}
\label{equation:2-1-2-3-2-5-first}
\delta_O(X)\geqslant \min\left\{\frac{1}{S_X(A)},\inf_{\substack{F/A\\ O\in C_A(F)}}\frac{A_A(F)}{S(W_{\bullet,\bullet}^A;F)}\right\}=\min\left\{\frac{16}{11},\inf_{\substack{F/A\\ O\in C_A(F)}}\frac{A_A(F)}{S(W_{\bullet,\bullet}^A;F)}\right\}
\end{equation}

where the infimum is taken by all prime divisors $F$ over the surface $A$ with ,
and $S(W_{\bullet,\bullet}^A;F)$ can be computed using <cit.> as follows:

S\big(W_{\bullet,\bullet}^A; F\big)=
\frac{3}{4d}\int_1^{2}\big(P(u)\big\vert_{A}\big)^2(u-1)\mathrm{ord}_{F}\big(E\big\vert_{A}\big)du+
\frac{3}{4d}\int_{0}^{2}\int_0^\infty \mathrm{vol}\big(P(u)\big\vert_{A}-vF\big)dvdu.

Now, let $F$ be any prime divisor over the surface $A$ such that $O\in C_A(F)$. Since

&-K_A\ \text{if $0\leqslant u\leqslant 1$}, \\
&(2-u)(-K_A)\ \text{if $1\leqslant u\leqslant 2$},
\endaligned
\right.

we have

\begin{multline*}
\quad\quad\quad\quad\quad S\big(W_{\bullet,\bullet}^A; F\big)=
\frac{3}{4d}\int_1^{2}d(2-u)^2(u-1)\mathrm{ord}_{F}\big(E\big\vert_{A}\big)du+\\
+\frac{3}{4d}\int_{0}^{1} \int_{0}^{\infty}\mathrm{vol}\big(-K_A-vF\big)dvdu+\frac{3}{4d}\int_{1}^{2}\int_{0}^{\infty}\mathrm{vol}\big((2-u)(-K_A)-vF\big)dvdu=\\
\end{multline*}

Therefore, if $O\not\in E$, then $\mathrm{ord}_{F}(E\vert_{A})=0$, which implies that

Similarly, if $O\in E$, then $\mathrm{ord}_{F}(E\vert_{A})\leqslant A_A(F)$, because $(A,E\vert_A)$ is log canonical,
so that

S\big(W_{\bullet,\bullet}^A; F\big)=\frac{\mathrm{ord}_{F}\big(E\big\vert_{A}\big)}{16}+\frac{15}{16}S_A(F)\leqslant\frac{A_A(F)}{16}+\frac{15A_{A}(F)}{16\delta_{P}(A)}=\frac{\delta_{P}(A)+15}{16\delta_{P}(A)}A_{A}(F).

Now, using (<ref>), we obtain the required inequality.

Suppose $X$ is not K-stable. Let us seek for a contradiction.
Using the valuative criterion for K-stability [14, 16],
we see that there exists a prime divisor $\mathbf{F}$ over $X$ such that

\beta(\mathbf{F})=A_X(\mathbf{F})-S_X(\mathbf{F})\leqslant 0,

where $A_X(\mathbf{F})$ is a log discrepancy of the divisor $\mathbf{F}$, and $S_X(\mathbf{F})$ is defined in [14] or .
Let $Z$ be the center of the divisor $\mathbf{F}$ on $X$.
Then $Z$ is not a surface .
We see that $Z$ is an irreducible curve or a point. Let $P$ be a point in $Z$.
Then $\delta_P(X)\leqslant 1$.

One has $P\not\in E$.

Let us compute $S_X(E)$. Note that $S_X(E)<1$ by <cit.>.
Fix $u\in\mathbb{R}_{\geqslant 0}$.
Then the divisor $-K_X-uE$ is pseudoeffective $\iff$ it is nef $\iff$ $u\leqslant 1$.
Thus, we have

Suppose that $P\in E$. Let us seek for a contradiction.

Note that $E\cong\mathcal{C}\times\mathbb{P}^1$.
Let $\mathbf{s}$ be a fiber of the projection $\phi\vert_{E}\colon E\to\mathbb{P}^1$ that contains $P$,
and let $\mathbf{f}$ be a fiber of the projection $\pi\vert_{E}\colon E\to\mathcal{C}$.
Fix $u\in[0,1]$ and take $v\in\mathbb{R}_{\geqslant 0}$. Then

-K_X-uE\big\vert_{E}-v\mathbf{s}\equiv (1+u-v)\mathbf{s}+d(1-u)\mathbf{f}.

Thus, the divisor $-K_X-uE\big\vert_{E}-v\mathbf{s}$ is pseudoeffective $\iff$ it is nef  $\iff$ $v\leqslant 1+v$.
Therefore, using  <cit.>, we get

S\left(W_{\bullet,\bullet}^{E};\mathbf{s}\right)=\frac{3}{4d} \int_{0}^{1} \int_{0}^{1+u} 2d(1-u)(1+u-v)dvdu=\frac{11}{16}.

Similarly, using <cit.>, we get

S\left(W_{\bullet,\bullet, \bullet}^{E,\mathbf{s}};P\right)=\frac{3}{4d}\int_{0}^{1} \int_{0}^{1+u}\big(d(1-u)\big)^{2}dvdu=\frac{5d}{16}.

Therefore, it follows from <cit.> that

1\geqslant\frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant \min\left\{\frac{1}{S_{X}(E)}, \frac{1}{S\left(W_{\bullet,\bullet}^E,\mathbf{s}\right)},
\frac{1}{S\left(W_{\bullet,\bullet,\bullet}^{E,\mathbf{s}};P\right)}\right\}=\min \left\{\frac{8}{3}, \frac{16}{11}, \frac{16}{5d}\right\}\geqslant\frac{16}{15}>1,

which is a contradiction.

Let $A$ be the fiber of the del Pezzo fibration $\phi$ such that $A$ passes through the point $P$.
Then $A$ is a del Pezzo surface of degree $d\in\{1,2,3\}$ that has at most isolated singularities.
In particular, we see that $A$ is normal. Applying Lemmas <ref> and <ref>, we obtain

One has $\delta_P(A)\leqslant\frac{15}{16}$.

Since $1\geqslant\frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\delta_P(X)$ , we get $\delta_P(A)\leqslant\frac{15}{16}$ by Lemmas <ref> and <ref>.

The surface $A$ is singular.

If $A$ is smooth, then $\delta_P(A)\geqslant\delta(A)\geqslant\frac{3}{2}$ <cit.>, which contradicts Corollary <ref>.

Let $\overline{S}$ be a general surface in $|H|$ that passes through  $\pi(P)$,
and let $S$ be the proper transform on $X$ of the surface $\overline{S}$. Then

* the surface $\overline{S}$ is a smooth del Pezzo surface of degree $d$,

* the surface $\overline{S}$ intersects the curve $\mathcal{C}$ transversally at $d$ points,

* the induced morphism $\pi\vert_{S}\colon S\to\overline{S}$ is a blow up of the points $\overline{S}\cap\mathcal{C}$.

Observe that  $\phi\vert_S\colon S\to \mathbb{P}^1$ is an elliptic fibration given by the pencil $|-K_S|$.
Set $C=A\big\vert_{S}$.
Then $C$ is a reduced curve of arithmetic genus $1$ in $|-K_S|$ that has at most $d$ components.
In particular, if $d=1$, then $C$ is irreducible.
Therefore, the following cases may happen:

* the curve $C$ is irreducible, and $C$ is smooth at $P$,

* the curve $C$ is irreducible, and $C$ has an ordinary node at $P$,

* the curve $C$ is irreducible, and $C$ has an ordinary cusp at $P$,

* the curve $C$ is reducible.

Fix $u\in\mathbb{R}_{\geqslant 0}$.
Then $-K_X-uS$ is nef $\iff$ $u\leqslant 1$ $\iff$ $-K_X-uS$ is pseudoeffective.
Using this, we see that

which also follows from <cit.>. Moreover, if $u\in[0,1]$, then

where $\mathbf{e}_1,\ldots,\mathbf{e}_d$ are exceptional curves of the blow up $\pi\vert_{S}\colon S\to\overline{S}$.

Suppose that $C$ is irreducible. Then $C$ is singular at the point $P$.

As in the proof of Lemma <ref>, it follows from <cit.> that

1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant \min\left\{\frac{1}{S_X(S)},\frac{1}{S(W_{\bullet,\bullet}^S;C)},
\frac{1}{S(W_{\bullet, \bullet,\bullet}^{S,C};P)}\right\},

where $S(W_{\bullet,\bullet}^S;C)$ and $S(W_{\bullet, \bullet,\bullet}^{S,C};P)$ are defined in <cit.>.
Since we know that , we see that $S(W_{\bullet,\bullet}^S;C)\geqslant 1$ or $S(W_{\bullet, \bullet,\bullet}^{S,C};P)\geqslant 1$.
Let us compute these numbers.

Let $P(u,v)$ be the positive part of the Zariski decomposition of $(-K_X-uS)\vert_{S}-vC$,
and let $N(u,v)$ be its negative part, where $u\in[0,1]$ and $v\in\mathbb{R}_{\geqslant 0}$. Since

we see that $(-K_X-uS)\vert_{S}-vC$ is pseudoeffective $\iff$ $v\leqslant 2-u$.
Moreover, we have

&(1-u)\sum_{i=1}^{d}\mathbf{e}_i+(2-u-v)C\ \text{if $0\leqslant v\leqslant 1$}, \\
&(2-u-v)\Big(C+\sum_{i=1}^{d}\mathbf{e}_i\Big)\ \text{if $1\leqslant v\leqslant 2-u$}, \\
\endaligned
\right.

&0\ \text{if $0\leqslant v\leqslant 1$}, \\
&(v-1)\sum_{i=1}^{d}\mathbf{e}_i\ \text{if $1\leqslant v\leqslant 2-u$}. \\
\endaligned
\right.

Thus, it follows from <cit.> that

\begin{multline*}
\quad \quad \quad \quad \quad\quad \quad \quad  S\big(W^S_{\bullet,\bullet};C\big)=\frac{3}{4d}\int_0^1\int_0^{2-u}P(u,v)^2dvdu=\\
\end{multline*}

Thus, we conclude that  $S(W_{\bullet, \bullet,\bullet}^{S,C};P)\geqslant 1$.

Since $P\not\in\mathbf{e}_1\cup\cdots\cup\mathbf{e}_d$ by Lemma <ref>, it follows from <cit.> that

\begin{multline*}
\quad \quad \quad \quad \quad \quad \quad \quad \quad S\big(W_{\bullet, \bullet,\bullet}^{S,C};P\big)=\frac{3}{4d}\int_0^1\int_0^{2-u}\big(P(u,v)\cdot C\big)^2dvdu=\\
=\frac{3}{4d}\int_0^1\int_0^1d^2(u-1)^2dvdu+\frac{3}{4d}\int_0^1\int_1^{2-u}d^2(2-u-v)^2dvdu=\frac{5d}{16}<1,\quad \quad \quad
\end{multline*}

which is a contradiction.

Now, let us show that $C$ is reducible for $d\in\{1,2\}$.

Suppose that $C$ is irreducible.
Then $d=3$ and $C$ has a cusp at $P$.

By Lemma <ref>, the curve $C$ is singular at the point $P$.

Now, let $\sigma\colon\widetilde{S}\to S$ be the blow up of the point $P$, let $\mathbf{f}$ be the $\sigma$-exceptional curve,
and let $\widetilde{\mathbf{e}}_1,\ldots,\widetilde{\mathbf{e}}_d,\widetilde{C}$ be the proper transforms on $\widetilde{S}$ of the curves $\mathbf{e}_1,\ldots,\mathbf{e}_d,C$, respectively.
Then the curve $\widetilde{C}$ is smooth, and it follows from <cit.> that

1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\min\left\{\inf_{O\in\mathbf{f}}\frac{1}{S(W_{\bullet,\bullet,\bullet}^{\widetilde{S},\mathbf{f}};O)},
\frac{2}{S(V_{\bullet,\bullet}^{S};\mathbf{f})},\frac{1}{S_X(S)}\right\},

where $S(W_{\bullet, \bullet,\bullet}^{\widetilde{S},\mathbf{f}};O)$ and $S(V_{\bullet,\bullet}^{S};\mathbf{f})$
are defined in <cit.>.
Since we know that $S_X(S)<1$, we see that $S(V_{\bullet,\bullet}^{S};\mathbf{f})\geqslant 2$ or
there exists a point $O\in\mathbf{f}$ such that $S(W_{\bullet,\bullet,\bullet}^{\widetilde{S},\mathbf{f}};O)\geqslant 1$.

Let us compute $S(V_{\bullet,\bullet}^{S};\mathbf{f})$.
Fix $u\in[0,1]$ and $v\in\mathbb{R}_{\geqslant 0}$.
Since $\sigma^*(C)\sim\widetilde{C}+2\mathbf{f}$, we get

\sigma^*\big((-K_X-uS)\vert_{S}\big)-v\mathbf{f}\sim_{\mathbb{R}}(2-u)\widetilde{C}+(4-2u-v)\mathbf{f}+(1-u)\sum_{i=1}^{d}\widetilde{\mathbf{e}}_i.

Then $\sigma^*((-K_X-uS)\vert_{S})-v\mathbf{f}$ is pseudoeffective $\iff$ $v\leqslant 4-2u$.

Let $P(u,v)$ be the positive part of the Zariski decomposition of $\sigma^*((-K_X-uS)\vert_{S})-v\mathbf{f}$,
and let $N(u,v)$ be its negative part.

&(2-u)\widetilde{C}+(4-2u-v)\mathbf{f}+(1-u)\sum_{i=1}^{d}\widetilde{\mathbf{e}}_i\ \text{if $0\leqslant v\leqslant \frac{d-du}{2}$}, \\
&\frac{8+d-4u-du-2v}{4}\widetilde{C}+(4-2u-v)\mathbf{f}+(1-u)\sum_{i=1}^{d}\widetilde{\mathbf{e}}_i\ \text{if $\frac{1-u}{2}\leqslant v\leqslant \frac{4+d-du}{2}$}, \\
&\frac{4-2u-v}{4-d}\Big(2\widetilde{C}+(4-d)\mathbf{f}+2\sum_{i=1}^{d}\widetilde{\mathbf{e}}_i\Big)\ \text{if $\frac{4+d-du}{2}\leqslant v\leqslant 4-2u$},
\endaligned
\right.

&0\ \text{if $0\leqslant v\leqslant \frac{d-du}{2}$}, \\
&\frac{2v+du-d}{4}\widetilde{C}\ \text{if $\frac{d-du}{2}\leqslant v\leqslant \frac{4+d-du}{2}$}, \\
&\frac{2v+du-2d}{4-d}\widetilde{C}+\frac{2v+du-4-d}{4-d}\sum_{i=1}^{d}\widetilde{\mathbf{e}}_i\ \text{if $\frac{4+d-du}{2}\leqslant v\leqslant 4-2u$}.
\endaligned
\right.

Thus, using <cit.>, we get

\begin{multline*}
\end{multline*}

Therefore, there exists a point $O\in\mathbf{f}$ such that $S(W_{\bullet,\bullet,\bullet}^{\widetilde{S},\mathbf{f}};O)\geqslant 1$.

Let us compute $S(W_{\bullet,\bullet,\bullet}^{\widetilde{S},\mathbf{f}};O)$. Observe that

&v\ \text{if $0\leqslant v\leqslant \frac{d-du}{2}$}, \\
&\frac{d-du}{2}\ \text{if $\frac{d-du}{2}\leqslant v\leqslant \frac{4+d-du}{2}$}, \\
&\frac{d(4-2u-v)}{4-d}\ \text{if $\frac{4+d-du}{2}\leqslant v\leqslant 4-2u$}.
\endaligned
\right.

Thus, it follows from <cit.> that

\begin{multline*}
\quad\quad\quad\quad S(W_{\bullet,\bullet,\bullet}^{\widetilde{S},\mathbf{f}};O)=\frac{3}{4d}\int_0^1\int_0^{4-2u}\Big(\big(P(u,v)\cdot\mathbf{f}\big)\Big)^2dvdu+\\
\frac{3}{4d}\int_0^1\int_{\frac{4+d-du}{2}}^{4-2u}\Big(\frac{d(4-2u-v)}{4-d}\Big)^2dvdu+\\
\end{multline*}

Therefore, if $O\not\in\widetilde{C}$, we obtain $S(W_{\bullet,\bullet,\bullet}^{\widetilde{S},\mathbf{f}};O)=\frac{5d}{32}$,
which contradicts to $S(W_{\bullet,\bullet,\bullet}^{\widetilde{S},\mathbf{f}};O)\geqslant 1$.
Similarly, if $O\in\widetilde{C}$ and $\widetilde{C}$ intersects the curve $\mathbf{f}$ transversally at the point $O$,

which again contradicts $S(W_{\bullet,\bullet,\bullet}^{\widetilde{S},\mathbf{f}};O)\geqslant 1$.
Therefore, the curves $\widetilde{C}$ and $\mathbf{f}$ are tangent at $O$,
which implies that $C$ has a cusp at the point $P$.

Thus, to proceed, we may assume that $d=1$ or $d=2$.

Now, let us consider the following commutative diagram:

\xymatrix{
\widetilde{S}\ar@{->}[d]_{\sigma}&&\widehat{S}\ar@{->}[ll]_{\rho}&&\overline{S}\ar@{->}[ll]_{\eta}\ar@{->}[d]^{\psi}\\%
S &&&& \mathscr{S}\ar@{->}[llll]_{\upsilon}}

where $\rho$ is the blow up of the point $\widetilde{C}\cap\mathbf{f}$,
the morphism $\eta$ is the blow up of the point in the $\rho$-exceptional curve
that is contained in the the proper transform of the curve $\widetilde{C}$,
the map $\psi$ is the contraction of the proper transforms of both $(\sigma\circ\rho)$-exceptional curves,
and $\upsilon$ is the birational contraction of the proper transform of the $\eta$-exceptional curve.
Let $\mathscr{F}$ be the $\upsilon$-exceptional curve, let $\mathscr{C}$ be the proper transform on $\mathscr{S}$ of the curve $C$.
Then $\mathscr{F}$ and $\mathscr{C}$ are smooth,
$\mathscr{C}^2=-6$, $\mathscr{F}^{2}=-\frac{1}{6}$, $\mathscr{C}\cdot\mathscr{F}=1$, and $\upsilon^*(C)=\mathscr{C}+6\mathscr{F}$.

Observe that $\mathscr{F}$ contains two singular points of the surfaces $\mathscr{S}$,
which are quotient singular points of type $\frac{1}{2}(1,1)$ and $\frac{1}{3}(1,1)$.
Denote these points by $Q_2$ and $Q_3$, respectively.
Note that  $\mathscr{C}$ does not contain $Q_2$ and $Q_3$.
Write $\Delta_{\mathscr{F}}=\frac{1}{2}Q_2+\frac{2}{3}Q_3$.
Then, since $A_S(\mathscr{F})=5$, it follows from <cit.> that

1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\min\left\{\inf_{Q\in\mathscr{F}}\frac{1-\mathrm{ord}_{Q}(\Delta_{\mathscr{F}})}{S(W_{\bullet,\bullet,\bullet}^{\mathscr{S},\mathscr{F}};Q)},
\frac{5}{S(V_{\bullet,\bullet}^{S};\mathscr{F})},\frac{1}{S_X(S)}\right\}.

But we already proved that $S_X(S)<1$. Hence, we conclude that $S(V_{\bullet,\bullet}^{S};\mathscr{F})\geqslant 5$ or
there exists a point $Q\in\mathscr{F}$ such that
$S(W_{\bullet,\bullet,\bullet}^{\mathscr{S},\mathscr{F}};Q)\geqslant 1-\mathrm{ord}_{Q}(\Delta_{\mathscr{F}})$.

Let us compute $S(V_{\bullet,\bullet}^{S};\mathscr{F})$. Take $v\in\mathbb{R}_{\geqslant 0}$. Then

\upsilon^*\big(P(u)|_S\big)-v\mathscr{F}\sim_{\mathbb{R}}(2-u)\mathscr{C}+(1-u)\sum_{i=1}^{d}\mathscr{E}_i+(12-6u-v)\mathscr{F},

where $\mathscr{E}_i$ is the proper transform on $\mathscr{S}$ of the $(-1)$-curve $\mathbf{e}_i$.
Using this, we conclude that the divisor $\upsilon^*(P(u)|_S)-v\mathscr{F}$ is pseudoeffective $\iff$ $v\leqslant 12-6u$.

Let $\mathscr{P}(u,v)$ be the positive part of the Zariski decomposition of $\upsilon^*(P(u)|_S)-v\mathscr{F}$,
and let $\mathscr{N}(u,v)$ be its negative part.

\mathscr{P}(u,v)=\left\{\aligned
&(2-u)\mathscr{C}+(1-u)\sum_{i=1}^{d}\mathscr{E}_i+(12-6u-v)\mathscr{F}\ \text{if $0\leqslant v\leqslant d(1-u)$}, \\
&\frac{12+d-(6+d)u-v}{6}\mathscr{C}+(1-u)\sum_{i=1}^{d}\mathscr{E}_i+(12-6u-v)\mathscr{F}\ \text{if $d(1-u)\leqslant v\leqslant 6+d-du$}, \\
&\frac{12-6u-v}{6-d}\Big(\mathscr{C}+\sum_{i=1}^{d}\mathscr{E}_i+(6-d)\mathscr{F}\Big)\ \text{if $6+d-du\leqslant v\leqslant 12-6u$},
\endaligned
\right.

\mathscr{N}(u,v)=\left\{\aligned
&0\ \text{if $0\leqslant v\leqslant d(1-u)$}, \\
&\frac{v+d(u-1)}{6}\mathscr{C}\ \text{if $d(1-u)\leqslant v\leqslant 6+d-du$}, \\
&\frac{v+du-2d}{6-d}\mathscr{C}+\frac{v+du-6-d}{6-d}\sum_{i=1}^{d}\mathscr{E}_i\ \text{if $6+d-du\leqslant v\leqslant 12-6u$}.
\endaligned
\right.

This gives

\mathscr{P}(u,v)^2=\left\{\aligned
&\frac{6du^2-24du-v^2+18d}{6}\ \text{if $0\leqslant v\leqslant d(1-u)$}, \\
&\frac{d(1-u)(18-(d+6)u+d-2v)}{6}\ \text{if $d(1-u)\leqslant v\leqslant 6+d-du$}, \\
&\frac{d(12-6u-v)^2}{6(6-d)}\ \text{if $6+d-du\leqslant v\leqslant 12-6u$}.
\endaligned
\right.

Thus, using <cit.> and integrating, we get

since $d=1$ or $d=2$.
Thus, there is a point $Q\in\mathscr{F}$ such that $S(W_{\bullet,\bullet,\bullet}^{\mathscr{S},\mathscr{F}};Q)\geqslant 1-\mathrm{ord}_{Q}(\Delta_{\mathscr{F}})$.

Now, using <cit.> again, we see that

\begin{multline*}
\quad \quad \quad \quad \quad S\big(W_{\bullet,\bullet,\bullet}^{\mathscr{S},\mathscr{F}};Q\big)=\frac{3}{4d}\int_0^1\int_0^{12-6u}\Big(\big(\mathscr{P}(u,v)\cdot\mathscr{F}\big)\Big)^2dvdu+\\
+\frac{6}{4d}\int_0^1\int_0^{12-6u}\big(P(u,v)\cdot\mathscr{F}\big)\mathrm{ord}_Q\big(\mathscr{N}(u,v)\big|_{\mathscr{F}}\big)dvdu.\quad \quad \quad \quad \quad
\end{multline*}

On the other hand, we have

\mathscr{P}(u,v)\cdot\mathscr{F}=\left\{\aligned
&\frac{v}{6}\ \text{if $0\leqslant v\leqslant d(1-u)$}, \\
&\frac{d(1-u)}{6}\ \text{if $d(1-u)\leqslant v\leqslant 6+d-du$}, \\
&\frac{d(12-6u-v)}{6(6-d)}\ \text{if $6+d-du\leqslant v\leqslant 12-6u$},
\endaligned
\right.

\mathscr{N}(u,v)\cdot\mathscr{F}=\left\{\aligned
&0\ \text{if $0\leqslant v\leqslant d(1-u)$}, \\
&\frac{v-d(1-u)}{6}\ \text{if $d(1-u)\leqslant v\leqslant 6+d-du$}, \\
&\frac{v+du-2d}{6-d}\ \text{if $6+d-du\leqslant v\leqslant 12-6u$}.
\endaligned
\right.

In particular, we have

Hence, if $Q\not\in\mathscr{C}$, then
$\frac{1}{3}\leqslant 1-\mathrm{ord}_{Q}(\Delta_{\mathscr{F}})\leqslant S(W_{\bullet,\bullet,\bullet}^{\mathscr{S},\mathscr{F}};Q)=\frac{5d}{96}<\frac{1}{3}$,
which is absurd.
Thus, we conclude that $Q=\mathscr{C}\cap\mathscr{F}$. Then

\begin{multline*}
\leqslant\frac{5d}{96}+\frac{6}{4d}\int_0^1\int_0^{12-6u}\big(P(u,v)\cdot\mathscr{F}\big)\big(\mathscr{N}(u,v)\cdot\mathscr{F}\big)dvdu=\frac{11}{16}<1,
\end{multline*}

which is a contradiction,
since $S(W_{\bullet,\bullet,\bullet}^{\mathscr{S},\mathscr{F}};Q)\geqslant 1-\mathrm{ord}_{Q}(\Delta_{\mathscr{F}})=1$.

In particular, we conclude that either $d=2$ or $d=3$.

All smooth Fano threefolds in the family №2.1 are K-stable.

Recall that $A$ is the fiber of the del Pezzo fibration $\phi\colon X\to\mathbb{P}^1$ that passes through $P$.
Note also that we have the following possibilities:

* $d=2$, and $A$ is a double cover of $\mathbb{P}^2$ branched over a reduced quartic curve;

* $d=3$, and $A$ is a normal cubic surface in $\mathbb{P}^3$.

Observe that $C=S\cap A$, where $S$ is a general surface in $|\pi^*(H)|$ that contains the point $P$.
Since $C$ is singular at $P$, the surface $A$ must be singular at $P$, which confirms Corollary <ref>.
Now, using classifications of reduced singular plane quartic curves and singular normal cubic surfaces [6],
we see that $P=\mathrm{Sing}(A)$, and one of the following three cases holds:

* $d=2$, and $A$ is a double cover of $\mathbb{P}^2$ branched
over $4$ lines intersecting in a point;

* $d=3$, and $A$ is a cone in $\mathbb{P}^3$ over a smooth plane cubic curve;

* $d=3$, and $A$ has Du Val singular point of type $\mathbb{D}_4$, $\mathbb{D}_5$, or $\mathbb{E}_6$.

Let us show that the first case is impossible.

One has $d=3$.

Suppose that $d=2$.
Then $P=\mathrm{Sing}(A)$, and $A$ is a double cover of $\mathbb{P}^2$ branched
over a reduced reducible plane quartic curve that is a union of $4$ distinct lines passing through one point.
Let us seek for a contradiction.

Let $\alpha\colon\widetilde{X}\to X$ be the blow up of the point $P$,
let $E_P$ be the $\alpha$-exceptional divisor, and let $\widetilde{A}$ be the proper transform on $\widetilde{X}$ of the surface $A$.
Then $\widetilde{A}\cap E_P$ is a line $L\subset E_P\cong\mathbb{P}^2$,
and the surface $\widetilde{A}$ is singular along this line.
Let $\beta\colon\overline{X}\to\widetilde{X}$ be the blow up of the line $L$,
let $E_L$ be the $\beta$-exceptional divisor,
let $\overline{A}$ be the proper transform on $\overline{X}$ of the surface $\widetilde{A}$,
and let $\overline{E}_P$ be the proper transforms on $\overline{X}$  of the surface $E_P$.

* $E_L\cong\mathbb{F}_2$,

* the intersection $\overline{E}_P\cap E_L$ is the $(-2)$-curve in $E_L$,

* the surface $\overline{A}$ is smooth, and the exists a $\mathbb{P}^1$-bundle $\overline{A}\to\mathcal{C}$,

* $\overline{A}\cap E_L$ is a smooth elliptic curve that is a section of the $\mathbb{P}^1$-bundle $\overline{A}\to\mathcal{C}$,

* the surfaces $\overline{A}$ and $\overline{E}_P$ are disjoint,

* $\overline{E}_P\cong\mathbb{P}^2$ and $\overline{E}_P\vert_{\overline{E}_P}\cong\mathcal{O}_{\mathbb{P}^2}(-2)$.

There is a birational contraction $\gamma\colon\overline{X}\to\widehat{X}$ of the surface $\overline{E}_P$ such that
$\widehat{X}$ is a projective threefold that has one singular point $O=\gamma(\overline{E}_P)$,
which is a terminal cyclic quotient singularities of type $\frac{1}{2}(1,1,1)$.
Thus, there exists the following commutative diagram

\overline{X}\ar[rr]^{\beta}\ar[d]_{\gamma}&&\widetilde{X}\ar[d]^{\alpha}\\
\widehat{X}\ar[rr]_{\sigma}&&X}

where $\sigma$ is a birational morphism that contracts the surface $\gamma(E_L)$ to the point $P$.

Let $G=\gamma(E_L)$, let $\widehat{A}=\gamma(\overline{A})$, and  let $\widehat{E}$ be the proper transform on $\widehat{X}$ of the surface $E$.
Then $A_{X}(G)=4$, $\sigma^{*}(-K_X)\sim 2\widehat{A}+\widehat{E}+8G$ and $\sigma^{*}(A)\sim\widehat{A}+4G$.

Note that $\widehat{A}\cong\overline{A}$ and $G\cong\mathbb{P}(1,1,2)$,
so we can identify $G$ with a quadric cone in $\mathbb{P}^3$.
Note also that $O$ is the vertex of the cone $G$. Moreover, by construction, we have $O\not\in\widehat{A}$.
Furthermore, the exists a $\mathbb{P}^1$-bundle $\widehat{A}\to\mathcal{C}$
such that $G\vert_{\widehat{A}}$ is its section.

Let $\mathbf{g}$ be a ruling of the quadric cone $G$, let $\mathbf{l}$ be a fiber of the $\mathbb{P}^1$-bundle $\widehat{A}\to\mathcal{C}$,
and let $\mathbf{f}$ be a fiber of the $\mathbb{P}^1$-bundle $\pi\circ\sigma\vert_{\widehat{E}}\colon\widehat{E}\to\mathcal{C}$.
Then $G\vert_{G}\sim_{\mathbb{Q}}-\mathbf{g}$ and $\widehat{A}\vert_{G}\sim_{\mathbb{Q}}4\mathbf{g}$.
Moreover, the intersections of the surfaces $G$, $\widehat{A}$, $\widehat{E}$ with  the curves $\mathbf{g}$, $\mathbf{l}$, $\mathbf{f}$
are given here:

$G$     $\widehat{A}$     $\widehat{E}$

$\mathbf{g}$     $-\frac{1}{2}$     2     0

$\mathbf{l}$     1     $-4$     1

$\mathbf{f}$     0     1     $-1$

Fix a non-negative real number $u$.
We have $\sigma^{*}(-K_X)-uG\sim_{\mathbb{R}} 2\widehat{A}+\widehat{E}+(8-u)G$,
which implies that  $\sigma^{*}(-K_X)-uG$ is pseudo-effective $\iff$ $u\in[0,8]$.
Furthermore, if $u\in[0,8]$, then the Zariski decomposition of the divisor $\sigma^{*}(-K_X)-uG$ can be described as follows:

&2\widehat{A}+\widehat{E}+(8-u)G\ \text{if $0\leqslant u\leqslant 1$}, \\
&\frac{9-u}{4}\widehat{A}+\widehat{E}+(8-u)G \ \text{if $1\leqslant u\leqslant 5$}, \\
&\frac{8-u}{3}\big(\widehat{A}+\widehat{E}+3G\big)\ \text{if $5\leqslant u\leqslant 8$}, \\
\endaligned
\right.

&0\ \text{if $0\leqslant u\leqslant 1$}, \\
&\frac{u-1}{4}\widehat{A}\ \text{if $1\leqslant u\leqslant 5$}, \\
&\frac{u-2}{3}\widehat{A}+\frac{u-5}{3}\widehat{E}\ \text{if $5\leqslant u\leqslant 8$}, \\
\endaligned
\right.

where $P(u)$ and $N(u)$ are the positive and the negative parts of the decomposition. Then

\mathrm{vol}\big(\sigma^{*}(-K_X)-uG\big)=P(u)^3=\left\{\aligned
&8-\frac{u^3}{8}\ \text{if $0\leqslant u\leqslant 1$}, \\
&\frac{18-3u}{2}\ \text{if $1\leqslant u\leqslant 5$}, \\
&\frac{(8-u)^{3}}{18}\ \text{if $5\leqslant u\leqslant 8$}. \\
\endaligned
\right.

Integrating, we get $S_{X}(G)=\frac{27}{8}<4=A_{X}(G)$.
But <cit.> gives

1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\min\left\{\frac{A_{X}(G)}{S_{X}(G)},
\inf_{Q\in G}\delta_{Q}\big(G,V^G_{\bullet,\bullet}\big)\right\}=\min\left\{\frac{32}{27},\inf_{Q\in G}\delta_{Q}\big(G,V^G_{\bullet,\bullet}\big)\right\},

where $\delta_{Q}(G,V^G_{\bullet,\bullet})$ is defined in [15].
Moreover, if $Q$ is a point in $G$ and $Z$ is a smooth curve in $G$ that passes through $Q$,
then it follows from <cit.> that

\delta_{Q}\big(G,V^G_{\bullet,\bullet}\big)
\geqslant\min\left\{\frac{1}{S\big(V^G_{\bullet,\bullet};Z\big)},\frac{1-\mathrm{ord}_Q(\Delta_Z)}{S\big(W^{G,Z}_{\bullet,\bullet,\bullet};Q\big)}\right\},

where $S(V^G_{\bullet,\bullet};Z)$ and $S(W^{G,Z}_{\bullet,\bullet,\bullet};Q)$ are defined in [15], and

\Delta_Z=\left\{\aligned
&0\ \text{if $O\not\in Z$}, \\
&\frac{1}{2}O\ \text{if $O\in Z$}.
\endaligned
\right.

Let us show that $\delta_{Q}\big(G,V^G_{\bullet,\bullet}\big)>1$ for every $Q\in G$, which would imply a contradiction.

Let $\mathscr{C}=\widehat{A}\vert_{G}$, and let $\ell$ be a curve in $|\mathbf{f}|$ that passes through $Q$.
Then $O\not\in\ell$ and $O\in\ell$, so that $\Delta_{\ell}=0$ and $\Delta_{\ell}=\frac{1}{2}O$.
Take $v\in\mathbb{R}_{\geqslant 0}$. Then

&(u-v)\mathbf{g}\ \text{if $0\leqslant u\leqslant 1$}, \\
&(1-v)\mathbf{g}\ \text{if $1\leqslant u\leqslant 5$}, \\
&\frac{8-u-3v}{3}\mathbf{g}\ \text{if $5\leqslant u\leqslant 8$}.\\
\endaligned
\right.

Now, using <cit.>, we get

\begin{multline*}
\frac{3}{8}\int_0^8\int_0^\infty \mathrm{vol}\big(P(u)\big\vert_{G}-v\ell\big)dvdu=\frac{3}{8}\int_{0}^{1}\int_{0}^{u}\frac{(u-v)^{2}}{2}dvdu+\\
\end{multline*}

Similarly, if $Q\not\in\mathscr{C}$, then it follows from <cit.> that

\begin{multline*}
\frac{3}{8}\int_{0}^{1}\int_{0}^{u}\Big(\big(P(u)\big\vert_{G}-v\ell\big)\cdot\ell\Big)^2dvdu+
\frac{3}{8}\int_{1}^{5}\int_{0}^{1}\Big(\big(P(u)\big\vert_{G}-v\ell\big)\cdot\ell\Big)^2dvdu+\\
\frac{3}{8}\int_{0}^{1}\int_{0}^{u}\frac{(u-v)^{2}}{4}dvdu+\\
\frac{3}{8}\int_{5}^{8}\int_{0}^{\frac{8-u}{3}}\frac{(8-u-3v)^2}{36}dvdu=\frac{5}{32}.
\end{multline*}

so that $S(W_{\bullet,\bullet,\bullet}^{G,\ell};Q)=\frac{5}{32}$,
which implies that $\delta_{Q}(G,V^G_{\bullet,\bullet})\geqslant\frac{16}{5}$. Likewise, if $Q\in\mathscr{C}$, then

\begin{multline*}
\frac{3}{8}\int_{0}^{8}\big(P(u)^{2}\cdot G\big)\cdot\mathrm{ord}_{\mathscr{C}}\Big(N(u)\big\vert_{G}\Big)du+
\frac{3}{8}\int_0^8\int_0^\infty \mathrm{vol}\big(P(u)\big\vert_{G}-v\mathscr{C}\big)dvdu=\\
+\frac{3}{8}\int_{1}^{5}\int_{0}^{\frac{1}{4}} \frac{(1-4v)^{2}}{2}dvdu+\frac{3}{8}\int_{5}^{8}\int_{0}^{\frac{8-u}{12}}\frac{(8-u-12v)^{2}}{18}dvdu=\frac{11}{16}
\end{multline*}

\begin{multline*}
\quad\quad\quad\quad S\big(W_{\bullet,\bullet,\bullet}^{G,\mathscr{C}};Q\big)=
\frac{3}{8}\int_{0}^{1}\int_{0}^{\frac{u}{4}}\Big(\big(P(u)\big\vert_{G}-v\mathscr{C}\big)\cdot\mathscr{C}\Big)^2dvdu+\\
=\frac{3}{8}\int_{0}^{1} \int_{0}^{\frac{u}{4}}(2u-8v)^{2}dvdu+\frac{3}{8}\int_{1}^{5} \int_{0}^{\frac{1}{4}}(2-8v)^{2}dvdu+\\
\end{multline*}

This implies that $\delta_{Q}(G,V^G_{\bullet,\bullet})\geqslant\min\left\{\frac{16}{11},\frac{8}{5}\right\}=\frac{16}{11}$,
which is a contradiction.

All smooth Fano threefolds in the family №2.3 are K-stable.

We see that $d=3$, so that $A$ is a singular cubic surface in $\mathbb{P}^3$ such that $P=\mathrm{Sing}(A)$.
Let $\sigma\colon\widehat{X}\to X$ be the blow up of the point $P$, and let $G$ be the $\sigma$-exceptional surface.
Denote by $\widehat{A}$ and $\widehat{E}$ the proper transforms on $\widehat{X}$ of the surfaces $A$ and $E$, respectively.

The surface $A$ has Du Val singularities.

Suppose that $A$ is a cone in $\mathbb{P}^3$ with vertex $P$. Take $u\in\mathbb{R}_{\geqslant 0}$. Then

\sigma^*(-K_X)-vG\sim_{\mathbb{R}}2\widehat{A}+\widehat{E}+(6-u)G.

Thus, the divisor $\sigma^*(-K_X)-vG$ is pseudo-effective $\iff$ $u\in[0,6]$.
Moreover, if $u\in[0,6]$, then the Zariski decomposition of the divisor $\sigma^*(-K_X)-vG$ can be described as follows:

&2\widehat{A}+E+(6-u)G\ \text{if $0\leqslant u\leqslant 1$}, \\
&\frac{7-u}{3}\widehat{A}+\widehat{E}+(6-u)G\ \text{if $1\leqslant u\leqslant 4$}, \\
&\frac{6-u}{2}(\widehat{A}+\widehat{E}+2G)\ \text{if $4\leqslant u\leqslant 6$}, \\
\endaligned
\right.

&0\ \text{if $0\leqslant u\leqslant 1$}, \\
&\frac{u-1}{3}\widehat{A}\ \text{if $1\leqslant u\leqslant 4$}, \\
&\frac{u-2}{2}\widehat{A}+\frac{u-4}{2}\widehat{E}\ \text{if $4\leqslant u\leqslant 6$}, \\
\endaligned
\right.

where $P(u)$ and $N(u)$ are the positive and the negative parts of the Zariski decomposition, respectively.
Using this, we compute

Let us apply <cit.>,  , <cit.>
using notations introduced in <cit.>. To start with, we apply <cit.> to get

\begin{equation}
\label{equation:2-5-d-3-cone}
1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\min\left\{\frac{A_{X}(G)}{S_{X}(G)},
\inf_{Q\in G}\delta_{Q}\big(G,V^G_{\bullet,\bullet}\big)\right\}=\min\left\{\frac{48}{43},\inf_{Q\in G}\delta_{Q}\big(G,V^G_{\bullet,\bullet}\big)\right\},
\end{equation}

where $\delta_{Q}(G,V^G_{\bullet,\bullet})$ is defined in <cit.>.
Let $Q$ be an arbitrary point in the surface $G$, and let $\ell$ is a general line in $G\cong\mathbb{P}^2$ that contains $Q$.
Then <cit.> gives

\delta_{Q}\big(G,V^G_{\bullet,\bullet}\big)
\geqslant\min\left\{\frac{1}{S\big(V^G_{\bullet,\bullet};\ell\big)},\frac{1}{S\big(W^{G,\ell}_{\bullet,\bullet,\bullet};Q\big)}\right\},

where $S(V^F_{\bullet,\bullet};\ell)$ and $S(W^{G,\ell}_{\bullet,\bullet,\bullet};Q)$ are defined in <cit.>.
Take $v\in\mathbb{R}_{\geqslant 0}$. Then

&(u-v)\ell\ \text{if $0\leqslant u\leqslant 1$}, \\
&(1-v)\ell\ \text{if $1\leqslant u\leqslant 4$}, \\
&\frac{6-u-2v}{2}\ell\ \text{if $4\leqslant u\leqslant 6$}.\\
\endaligned
\right.

Let $\mathscr{C}=\widehat{A}\vert_{G}$. Then $\mathscr{C}$ is a smooth cubic curve in $G\cong\mathbb{P}^2$. Let

&0\ \text{if $0\leqslant u\leqslant 1$}, \\
&\frac{u-1}{3}\mathscr{C}\ \text{if $1\leqslant u\leqslant 4$}, \\
&\frac{u-2}{2}\mathscr{C}\ \text{if $4\leqslant u\leqslant 6$}.
\endaligned
\right.

Now, using <cit.>, we get

\begin{multline*}
S\big(W^{G}_{\bullet,\bullet};\ell\big)=\frac{3}{12}\int_0^6\int_0^\infty \mathrm{vol}\big(P(u)\big\vert_{G}-v\ell\big)dvdu=\frac{3}{12}\int_{0}^{1}\int_{0}^{u}(u-v)^{2}dvdu+\\
\end{multline*}

Similarly, it follows from <cit.> that

\begin{multline*}
\frac{3}{12}\int_{0}^{1}\int_{0}^{u}\Big(\big(P(u)\big\vert_{G}-v\ell\big)\cdot\ell\Big)^2dvdu+
\frac{3}{12}\int_{1}^{4}\int_{0}^{1}\Big(\big(P(u)\big\vert_{G}-v\ell\big)\cdot\ell\Big)^2dvdu+\\
\end{multline*}

\begin{multline*}
+\frac{6}{12}\int_{4}^{6} \int_{0}^{\frac{6-u}{2}}\Big(\big(P(u)\big\vert_{G}-v\ell\big)\cdot\ell\Big)\mathrm{ord}_Q\big(N^\prime(u)\big|_\ell\big)dvdu\leqslant\\
\leqslant\frac{6}{12}\int_{1}^{4} \int_{0}^{1}\frac{(1-v)(u-1)}{3}dvdu+
\frac{6}{12}\int_{4}^{6} \int_{0}^{\frac{6-u}{2}}\frac{(6-u-2v)(u-2)}{4}dvdu=\frac{7}{12}.
\end{multline*}

So, we have $S(W_{\bullet,\bullet,\bullet}^{G,\ell};Q)\leqslant\frac{43}{48}$.
Then $\delta_{Q}(G,V^G_{\bullet,\bullet})>1$, which contradicts (<ref>).

Thus, we see that $P$ is a Du Val singular point of the surface $A$ of type $\mathbb{D}_4$, $\mathbb{D}_5$, $\mathbb{E}_6$.
Now, arguing as in the proof of <cit.>, we see that $\beta(\mathbf{F})>0$ if

* the inequality $\beta(G)>0$ holds,

* and for every prime divisor $\mathbf{E}$ over $X$ such that $C_X(\mathbf{E})$ is a curve containing $P$, the following inequality holds:

\frac{A_X(\mathbf{E})}{S_X(\mathbf{E})}\geqslant\frac{4}{3}.

Since $\beta(\mathbf{F})\leqslant 0$ by our assumption, we see that at least one of these conditions must fail.

One has $\beta(G)\geqslant\frac{465}{2048}$.

Let $\widehat{A}$ and $\widehat{E}$ be the proper transforms on $\widehat{X}$ of the surfaces $A$ and $E$, respectively.
Take $u\in\mathbb{R}_{\geqslant u}$. Then

\sigma^*(-K_{X})-uG\sim\sigma^*(2H-E)-uG\sim\sigma^*(2A+E)-uG\sim 2\widehat{A}+\widehat{E}+(4-u)G,

which easily implies that the divisor $-K_{\widehat{X}}-G$ is pseudoeffective $\iff$ $u\leqslant 4$,
because we can contract the surfaces $\widehat{A}$ and $\widehat{E}$ simultaneously after flops. Then

\beta(G)=A_X(G)-S_X(G)=3-\frac{1}{12}\int_0^4\mathrm{vol}\big(\sigma^*(-K_{X})-uG\big)du.

Note that $\sigma^*(-K_{X})-uG$ is nef for $u\in[0,1]$, because the divisor $-K_X$ is very ample.
Thus, if $u\in[0,1]$, then

\mathrm{vol}\big(\sigma^*(-K_{X})-uG\big)=\big(\sigma^*(-K_{X})-uG\big)^3=12-u^3.

Similarly, if $1\leqslant u\leqslant\frac{3}{2}$, then

\mathrm{vol}\big(\sigma^*(-K_{X})-uG\big)\leqslant\leqslant\mathrm{vol}\big(\sigma^*(-K_{X})-G\big)=\big(\sigma^*(-K_{X})-G\big)^3=11.

Finally, let us estimate $\mathrm{vol}(\sigma^*(-K_{X})-uG)$ in the case when $4\geqslant u>\frac{3}{2}$.

Let $Z$ be a general hyperplane section of the cubic surface $A$ that passes through $P$,
and let $\widehat{Z}$ be its proper transform on the threefold $\widehat{X}$.
Then $Z$ is an irreducible cuspidal cubic curve, and $\widehat{Z}\subset\widehat{A}$.
Observe that $(\sigma^*(-K_{X})-uG)\cdot\widehat{Z}=3-2u$ and $\widehat{A}\cdot\widehat{Z}=-4$,
so $\widehat{A}$ is contained in the asymptotic base locus of the divisor $\sigma^*(-K_{X})-uG$ for $u>\frac{3}{2}$.
Moreover, if $\sigma^*(-K_{X})-uG\sim_{\mathbb{R}}\widehat{D}+\lambda \widehat{A}$ for $\lambda\in\mathbb{R}_{\geqslant 0}$
and an effective $\mathbb{R}$-divisor $\widehat{D}$ whose support does not contain $\widehat{A}$,
then $\widehat{Z}\not\subset\widehat{D}$, which implies that

0\leqslant \widehat{D}\cdot\widehat{Z}=\Big(\sigma^*(-K_{X})-uG-\lambda \widehat{A}\Big)\cdot\widehat{Z}=3-2u-4\lambda,

so that $\lambda\geqslant\frac{3-2u}{4}$. Thus, if $4\geqslant u>\frac{3}{2}$, then

\mathrm{vol}\big(\sigma^*(-K_{X})-uG\big)\leqslant\mathrm{vol}\Big(\sigma^*(2H-E)-uG-\frac{2u-3}{4}\widehat{A}\Big).

Moreover, if $4\geqslant u>\frac{3}{2}$, then

\sigma^*(2H-E)-uG-\frac{2u-3}{4}\widehat{A}\sim_{\mathbb{R}}\frac{11-2u}{4}\sigma^{*}(H)-\frac{7-2u}{2}\sigma^*(E)-\frac{3}{2}G.

Therefore, if $4\geqslant u>\frac{3}{2}$, then

\mathrm{vol}\big(\sigma^*(-K_{X})-uG\big)\leqslant\mathrm{vol}\Big(\frac{11-2u}{4}\sigma^{*}(H)-\frac{7-2u}{2}\sigma^*(E)\Big)=\mathrm{vol}\Big(\frac{11-2u}{4}H-\frac{7-2u}{2}E\Big).

Furthermore, if $\frac{7}{2}\geqslant u>\frac{3}{2}$, then $\frac{11-2u}{4}H-\frac{7-2u}{2}E$ is nef, so that

\mathrm{vol}\Big(\frac{11-2u}{4}H-\frac{7-2u}{2}E\Big)=\Big(\frac{11-2u}{4}H-\frac{7-2u}{2}E\Big)^3=\frac{25-6u}{16}.

Similarly, if $4\geqslant u>\frac{7}{2}$, then

\mathrm{vol}\Big(\frac{11-2u}{4}H-\frac{7-2u}{2}E\Big)=\Big(\frac{11-2u}{4}H\Big)^3=\frac{(11-2u)^3}{256}.

Now, we can estimate $\beta(G)$ as follows

\begin{multline*}
\beta(G)=3-\frac{1}{12}\int_0^4\mathrm{vol}\big(\sigma^*(-K_{X})-uG\big)du\geqslant
\end{multline*}

as claimed.

Therefore, there exists a prime divisor $\mathbf{E}$ over $X$ such that $C_X(\mathbf{E})$ is a curve, $P\in C_X(\mathbf{E})$, and
$A_X(\mathbf{E})<\frac{4}{3}S_X(\mathbf{E})$. Set $Z=C_X(\mathbf{E})$. Then $\delta_O(X)<\frac{4}{3}$ for every point $O\in Z$.

One has $Z\subset A$, and $Z$ is a line in the cubic surface $A$.

Let $O$ be a general point in $Z$, and let $A_O$ be the fiber of $\phi$ that passes through $O$.
If $Z\not\subset A$, then $A_O$ is smooth, so that $\delta_O(A_O)\geqslant\frac{3}{2}$ by <cit.>,
which gives

\frac{4}{3}>\frac{A_X(\mathbf{E})}{S_X(\mathbf{E})}\geqslant\delta_O(X)\geqslant\mathrm{min}\Bigg\{\frac{16}{11},\frac{16\delta_O(A_O)}{16\delta_O(A_O)+15}\Bigg\}\geqslant\mathrm{min}\Bigg\{\frac{16}{11},\frac{16\times\frac{3}{2}}{\frac{3}{2}+15}\Bigg\}=\frac{16}{11}>\frac{4}{3},

by Lemma <ref>. This shows that $Z\subset A$ and $A_O=A$.

To complete the proof of the lemma, we have to show that $Z$ is a line in the surface $A$.
Suppose that $Z$ is not a line. Then the point $O$ is not contained in a line in the surface $A$, because $A$ contains finitely many lines [6].
Now, arguing as in the proof of <cit.>,
we get  $\delta_O(A)\geqslant\frac{3}{2}$.
So, applying Lemma <ref> again, we get a contradiction as above.

Now, our Auxiliary Theorem follows from the following lemma:

The surface $A$ does not have a singular point of type $\mathbb{D}_4$.

Suppose $A$ has singularity of type $\mathbb{D}_4$.
Then, it follows from [6] that,
for a suitable choice of coordinates $x$, $y$, $z$, $t$ on the projective space $\mathbb{P}^3$,
one of the following cases hold:

(A) $A=\{tx^2=y^3-z^3\}\subset \mathbb{P}^3$,

(B) $A=\{tx^2=y^3-z^3+xyz\}\subset \mathbb{P}^3$.

Note that $P=[0:0:0:1]$, and $A$ contains $6$ lines  [6].
In case (A), these lines are

\begin{align*}
\end{align*}

where $\omega_3$ is a primitive cube root of unity. In case (B), these lines are

\begin{align*}
\end{align*}

Note that $P=L_1\cap L_3\cap L_3$, $P\not\in L_4\cup L_5\cup L_6$ and
$-K_A\sim 2L_1+L_4\sim 2L_2+L_5\sim 2L_3+L_6$.

By Lemma <ref>, we may assume that $Z=L_1$.

Recall that $S_X(A)=\frac{11}{16}$, see the proof of Lemma <ref>.
Using <cit.>, we get

\frac{4}{3}>\frac{A_X(\mathbf{E})}{S_X(\mathbf{E})}\geqslant \min\left\{\frac{1}{S_X(A)},\frac{1}{S(W_{\bullet,\bullet}^A;L_1)}\right\}=\min\left\{\frac{16}{11},\frac{1}{S(W_{\bullet,\bullet}^A;L_1)}\right\},

where $S(W_{\bullet,\bullet}^A;L_1)$ is defined in <cit.>. Therefore, we conclude that $S(W_{\bullet,\bullet}^A;L_1)<\frac{4}{3}$.
Let us compute $S(W_{\bullet,\bullet}^A;L_1)$ using <cit.>.

To do this, we use notations introduced in the proof of Lemma <ref> applied to $O=P$.
Then, using <cit.> and computations from the proof of Lemma <ref>, we get

S\big(W_{\bullet,\bullet}^A; L_1\big)=\frac{1}{4}\int_{0}^{1}\int_0^\infty \mathrm{vol}\big(-K_A-vL_1\big)dvdu+\frac{1}{4}\int_{1}^{2}\int_0^\infty \mathrm{vol}\big((2-u)(-K_A)-vL_1\big)dvdu,

since $L_1\not\subset\mathrm{Supp}(N(u))$, since $L_1\not\subset E$. Let us compute $S(W_{\bullet,\bullet}^A; L_1)$. Take $v\in\mathbb{R}_{\geqslant 0}$.

Thus, the divisor $-K_A-vL_1$ is pseudoeffective $\iff$ $v\leqslant 2$, since $L_4^2=-1$. Fix $v\in[0,2]$.
Let $P(u,v)$ be the positive part of the Zariski decomposition of the divisor $-K_A-vL_1$,
and let $N(u,v)$ be its negative part. Then

&(2-v)L_1+L_4\ \text{if $0\leqslant v\leqslant 1$}, \\
&(2-v)(L_1+L_4)\ \text{if $1\leqslant v\leqslant 2$}, \\
\endaligned
\right.

&0\ \text{if $0\leqslant v\leqslant 1$}, \\
&(v-1)L_4\ \text{if $1\leqslant v\leqslant 2$}.
\endaligned
\right.

Thus, if $0\leqslant v\leqslant 1$, then $\mathrm{vol}(-K_A-vL_1)=3-2v$, because $L_1^2=0$ and $L_1\cdot L_4=0$.
Similarly, if $1\leqslant v\leqslant 2$, then $\mathrm{vol}(-K_A-vL_1)=(v-2)^2$. This gives

\frac{1}{4}\int_{0}^{1}\int_0^\infty \mathrm{vol}\big(-K_A-vL_1\big)dvdu=
\frac{1}{4}\int_{0}^{1}\int_0^1(3-2v)dvdu+\frac{1}{4}\int_{0}^{1}\int_1^2(v-2)^2dvdu=\frac{7}{12}

\begin{multline*}
\frac{1}{4}\int_{1}^{2}\int_0^\infty \mathrm{vol}\big((2-u)(-K_A)-vL_1\big)dvdu=
\frac{1}{4}\int_{1}^{2}\int_0^\infty(2-u)^3\mathrm{vol}\big((-K_A)-vL_1\big)dvdu=\\
\end{multline*}

Combining, we get $S(W_{\bullet,\bullet}^A;L_1)=\frac{35}{48}<\frac{3}{4}$. This is a contradiction.

§ FAMILY №2.2.

Let $R$ be a smooth surface of degree $(2,4)$ in $\mathbb{P}^1\times\mathbb{P}^2$,
let $\pi\colon X\to \mathbb{P}^1\times\mathbb{P}^2$ be a double cover ramified over surface $R$.
Then $X$ is a Fano threefold in the deformation family № 2.2.
Moreover, all smooth Fano threefolds in this family can be obtained this way.

Let $\mathrm{pr}_1\colon\mathbb{P}^1\times\mathbb{P}^2\to\mathbb{P}^1$
and $\mathrm{pr}_2\colon\mathbb{P}^1\times\mathbb{P}^2\to\mathbb{P}^2$ be the projections
to the first and the second factors, respectively.
Set $p_1=\mathrm{pr}_1\circ\pi$ and $p_2=\mathrm{pr}_2\circ\pi$.
We have the following commutative diagram:

$$\xymatrix{& X\ar[d]^{\pi}\ar<EMAIL_ADDRESS>\mathbb{P}^1 && \mathbb{P}^2}

where $p_1$ is a fibration into del Pezzo surfaces of degree $2$,
and $p_2$ is a conic bundle.

Let $S$ be a fiber of the morphism $p_1$. Then $S$ is irreducible and normal.

Left to the reader.

Let $S$ be a fiber of the morphism $p_1$, let $C$ be a fiber of the morphism $p_2$,
and let $P$ be a point in $S\cap C$.
Then $S$ or $C$ is smooth at $P$.

Local computations.

Now, we are ready to prove that $X$ is K-stable.
Recall from [9] that $\mathrm{Aut}(X)$ is finite.
Thus, the threefold $X$ is K-stable if and only if it is K-polystable [19].

Let $\tau$ be the Galois involution of the double cover $\pi\colon X\to\mathbb{P}^1\times\mathbb{P}^2$, and let $G=\langle\tau\rangle$.
Suppose that $X$ is not $K$-polystable.
Then it follows from <cit.> that there exists a $G$-invariant prime divisor $\mathbf{F}$ over $X$ such that

\beta(\mathbf{F})=A_X(\mathbf{F})-S_X(\mathbf{F})\leqslant 0.

Let $Z$ be the center of this divisor on $X$. Then $Z$ is not  a surface by <cit.>.
Hence, we see that either $Z$ is a $G$-invariant irreducible curve, or $Z$ is a $G$-fixed point.
Let us seek for a contradiction.

Let $P$ be a general point in $Z$, and let $S$ be the fiber of $p_1$ that passes through $P$.

The surface $S$ is singular at $P$.

Suppose that $S$ is smooth at $P$.
Let $C$ be a general curve in $|-K_S|$ that contains $P$.
Then $C$ is smooth.
Applying <cit.> to the flag $P\in C\subset S$, we get

1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant \min\left\{\frac{1}{S_X(S)},\frac{1}{S(W_{\bullet,\bullet}^S;C)},\frac{1}{S(W_{\bullet, \bullet,\bullet}^{S,C};P)}\right\}.

Since  by <cit.>,
we see that $S(W_{\bullet,\bullet}^S;C)\geqslant 1$ or $S(W_{\bullet, \bullet,\bullet}^{S,C};P)\geqslant 1$.
We refer the reader to <cit.> for definitions of $S_X(S)$, $S(W_{\bullet,\bullet}^S;C)$, $S(W_{\bullet, \bullet,\bullet}^{S,C};P)$.

Note that <cit.> requires $S$ to have Du Val singularities, but $S$ may have non-Du Val singularities.
Nevertheless, we still can apply <cit.> here, since the proof of <cit.>
remains valid in our case, because $S$ is smooth along $C$.

Let us compute $S(W_{\bullet,\bullet}^S;C)$ and $S(W_{\bullet, \bullet,\bullet}^{S,C};P)$.
Take $u\in\mathbb{R}_{\geqslant 0}$ and $v\in\mathbb{R}_{\geqslant 0}$.

$-K_X-uS$ is nef $\iff$ $-K_X-uS$ is pseudoeffective $\iff$ $u\leqslant 1$,

Similarly, if $u\in[0,1]$, then $(-K_X-uS)\vert_{S}-vC\sim_{\mathbb{R}}(1-v)(-K_S)$, so

$(-K_X-uS)\vert_{S}-vC$ is nef $\iff$ $(-K_X-uS)\vert_{S}-vC$ is pseudoeffective $\iff$ $v\leqslant 1$.

Now, applying <cit.>, we get

Similarly, using <cit.>, we get

S(W_{\bullet,\bullet,\bullet}^{S,C},P)=\frac{3}{6}\int_0^1\int_0^1\big((1-v)(-K_S)\cdot C\big)^2dvdu=\frac{2}{3}<1.

But we already know that $S(W_{\bullet,\bullet}^S;C)\geqslant 1$ or $S(W_{\bullet, \bullet,\bullet}^{S,C};P)\geqslant 1$.
This is a contradiction.

If $Z$ is a curve, then $S$ is smooth at $P$ by Lemma <ref>,
because $P$ is a general point in $Z$.
Hence, we conclude that $Z=P$, because $S$ is singular at the point $P$ by Lemma <ref>.
Recall that $Z$ is $G$-invariant. This implies that $\tau(P)\in R$.

Let $C$ be the fiber of $p_2$ that passes through $P$.
Then $C$ is smooth at $P$ by Lemma <ref>, because $S$ is singular at $P$.
Since $\tau(P)\in R$, we see that $C$ is irreducible and smooth.

Let $T$ be a sufficiently general surface in linear system $|\pi_2^*(\mathcal{O}_{\mathbb{P}^2}(1))|$ that contains $C$.
Since $C$ is smooth, it follows from Bertini's theorem that the surface $T$ is smooth.

As in the proof of Lemma <ref>,
it follows from <cit.> that

1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\min\left\{\frac{1}{S_X(T)},\frac{1}{S(W_{\bullet,\bullet}^T;C)},\frac{1}{S(W_{\bullet, \bullet,\bullet}^{T,C};P)}\right\}.

Moreover, it follows from <cit.> that .
Thus, we conclude that

\max\big\{S(W_{\bullet,\bullet}^T;C),S(W_{\bullet, \bullet,\bullet}^{T,C};P)\big\}\geqslant 1.

In fact, since $P$ is the center of the divisor $F$ on $X$,
<cit.> gives

\begin{equation}
\label{equation:2-2-strict}
\max\big\{S(W_{\bullet,\bullet}^T;C),S(W_{\bullet, \bullet,\bullet}^{T,C};P)\big\}>1.
\end{equation}

Now, let us compute $S(W_{\bullet,\bullet}^T;C)$ and $S(W_{\bullet, \bullet,\bullet}^{T,C};P)$ using the results obtained in <cit.>.

Take $u\in\mathbb{R}_{\geqslant 0}$ and $v\in\mathbb{R}_{\geqslant 0}$. Then

$-K_X-uT$ is nef $\iff$ $-K_X-uT$ is pseudoeffective $\iff$ $u\leqslant 1$,

Similarly, if $u\in[0,1]$, then

$(-K_X-uT)\vert_{T}-vC$ is nef $\iff$ $(-K_X-uT)\vert_{T}-vC$ is pseudoeffective $\iff$ $v\leqslant 1-u$,

because  $(-K_X-uT)\vert_{T}-vC\sim_{\mathbb{R}}S|_T+(1-u-v)C$.
So, using <cit.>, we get

Hence, it follows from (<ref>) that $S(W_{\bullet,\bullet,\bullet}^{T,C},P)>1$.
Now, using <cit.>, we get

S(W_{\bullet,\bullet,\bullet}^{T,C},P)=\frac{3}{6}\int_0^1\int_0^{1-u}\Big(\big(S|_T+(1-u-v)C\big)\cdot C\Big)^2dvdu=\frac{3}{6}\int_0^1\int_0^{1-u}4\,dvdu=1,

which is a contradiction. This shows that $X$ is K-stable.

All smooth Fano threefolds in the family №2.2 are K-stable.

§ FAMILY №2.4.

Let $\mathscr{S}$ and $\mathscr{S}^\prime$ be smooth cubic surfaces in $\mathbb{P}^3$ such that their intersection is a smooth curve of genus $10$.
Set $\mathscr{C}=\mathscr{S}\cap \mathscr{S}^\prime$, and let $\pi\colon X\to \mathbb{P}^3$ be the blow up of the curve $\mathscr{C}$.
Then $X$ is a smooth Fano threefold in the family №2.4,
and every smooth Fano threefold in this family can be obtained in this way.
Moreover, there exists a commutative diagram

\xymatrix{
\mathbb{P}^3\ar@{-->}[rr]&&\mathbb{P}^1}

where $\mathbb{P}^3\dasharrow\mathbb{P}^1$ is a map that is given by the pencil generated by the surfaces $\mathscr{S}$ and $\mathscr{S}^\prime$,
and $\phi$ is a fibration into cubic surfaces. Note that $-K_X^3=10$ and $\mathrm{Aut}(X)$ is finite [9].

Let $H=\pi^*(\mathcal{O}_{\mathbb{P}^3}(1))$, and let $E$ be the $\pi$-exceptional surface.
Then $-K_X\sim 4H-E$, the morphism $\phi$ is given by the linear system $|3H-E|$, and $E\cong\mathscr{S}\times\mathbb{P}^1$.

The goal of this section is to prove that $X$ is K-stable.
Suppose that $X$ is not K-stable. Let us seek for a contradiction.
First, using the valuative criterion for K-stability [14, 16],
we see that there exists a prime divisor $\mathbf{F}$ over $X$ such that

\beta(\mathbf{F})=A_X(\mathbf{F})-S_X(\mathbf{F})\leqslant 0.

Let $Z$ be the center of the divisor $\mathbf{F}$ on $X$.
Then $Z$ is not a surface by .
Therefore, either $Z$ is an irreducible curve or $Z$ is a point.
Fix a point $P\in Z$.

Let $A$ be the surface in $|3H-E|$ that contains $P$.
Fix $u\in\mathbb{R}_{\geqslant 0}$.
Let $\mathscr{P}(u)$ be the positive part of the Zariski decomposition of $-K_X-uA$,
and let $\mathscr{N}(u)$ be its negative part. Then

This implies that $-K_X-uA$ is pseudoeffective $\iff$ $u\leqslant\frac{4}{3}$.
Moreover, we have

\mathscr{P}(u)=\left\{\aligned
&(4-3u)H-(1-u)E\ \text{if $0\leqslant u\leqslant 1$}, \\
&(4-3u)H\ \text{if $1\leqslant v\leqslant \frac{4}{3}$}, \\
\endaligned
\right.

\mathscr{N}(u)=\left\{\aligned
&0\ \text{if $0\leqslant u\leqslant 1$}, \\
&(u-1)E\ \text{if $1\leqslant u\leqslant \frac{4}{3}$}. \\
\endaligned
\right.

Integrating, we obtain $S_X(A)=\frac{67}{120}<1$, which also follows from .

Note that $\pi(A)$ is a normal cubic surface in $\mathbb{P}^3$, and $\pi(A)$ is smooth along the curve $\mathscr{C}$.
In particular, we see that $A\cong\pi(A)$, and $A$ is smooth along the intersection $E\cap A$.

The surface $A$ is singular at the point $P$.

Suppose that $A$ is smooth at $P$.
Let $C$ be a general curve in $|-K_A|$ that passes through the point $P$.
Then $C$ is a smooth irreducible elliptic curve. Take $v\in\mathbb{R}_{\geqslant 0}$.

\mathscr{P}(u)\vert_{S}-vC\sim_{\mathbb{R}}\left\{\aligned
&(1-v)C\ \text{if $0\leqslant u\leqslant 1$}, \\
&(4-3u-v)C\ \text{if $1\leqslant u\leqslant \frac{4}{3}$}.
\endaligned
\right.

Therefore, using <cit.>, we obtain

Similarly, using <cit.>, we obtain

\begin{multline*}
\frac{3}{10}\int_0^1\int_0^1\big(3(1-v)\big)^2dvdu+\frac{3}{10}\int_1^{\frac{4}{3}}\int_0^{4-3u}\big(3(4-3u-v)\big)^2dvdu+\\
+\underbrace{\frac{6}{10}\int_1^{\frac{4}{3}}\int_0^{4-3u}3(4-3u-v)(u-1)dvdu}_{\text{if } P\in E}=\frac{39}{40}+\frac{1}{120}=\frac{59}{60}.
\end{multline*}

Therefore, it follows from <cit.> that

1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\min\left\{\frac{1}{S_X(A)},\frac{1}{S(W_{\bullet,\bullet}^A;C)},\frac{1}{S(W_{\bullet, \bullet,\bullet}^{A,C};P)}\right\}
\geqslant\min\left\{\frac{120}{67},\frac{40}{13},\frac{60}{59}\right\}>1,

which is absurd.

The point $P$ is not contained in the surface $E$.

Since $A\cong\pi(A)$, we may consider $A$ as a cubic surface in $\mathbb{P}^3$.

* either $\mathrm{mult}_P(A)=2$ and $A$ has Du Val singularities.

* or $\mathrm{mult}_P(A)=3$ and $A$ is a cone over a plane smooth cubic curve with vertex $P$.

One has $\mathrm{mult}_P(A)\ne 3$.

Let $\sigma\colon\widehat{X}\to X$ be a blow up of the point $P$, and let $G$ be the $\sigma$-exceptional surface.
Denote by $\widehat{A}$ and $\widehat{E}$ the proper transforms on $\widehat{X}$ of the surfaces $A$ and $E$, respectively.
Suppose that $\mathrm{mult}_P(A)=3$.  Take $u\in\mathbb{R}_{\geqslant 0}$. Then

\sigma^*(-K_X)-vG\sim_{\mathbb{R}}\frac{1}{3}\widehat{E}+\frac{4}{3}\widehat{A}+(4-u)G.

Thus, the divisor $\sigma^*(-K_X)-vG$ is pseudo-effective $\iff$ $u\in[0,4]$.
Moreover, if $u\in[0,4]$, then the Zariski decomposition of the divisor $\sigma^*(-K_X)-vG$ can be described as follows:

&\frac{1}{3}\widehat{E}+\frac{4}{3}\widehat{A}+(4-u)G\ \text{if $0\leqslant u\leqslant 1$}, \\
&\frac{1}{3}\widehat{E}+\frac{5-u}{3}\widehat{A}+(4-u)G \text{if $1\leqslant u\leqslant 4$}, \\
\endaligned
\right.

&0\ \text{if $0\leqslant u\leqslant 1$}, \\
&\frac{u-1}{3}\widehat{A}\ \text{if $1\leqslant u\leqslant 4$}, \\
\endaligned
\right.

where $P(u)$ and $N(u)$ are the positive and the negative parts of the Zariski decomposition, respectively.
Using this, we compute

As in the proof of Lemma <ref>,
let us use results obtained in <cit.> to get a contradiction.
Namely, applying <cit.>, we get

1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\min\left\{\frac{A_{X}(G)}{S_{X}(G)},
\inf_{Q\in G}\delta_{Q}\big(G,V^G_{\bullet,\bullet}\big)\right\}=
\min\left\{\frac{40}{31},\inf_{Q\in G}\delta_{Q}\big(G,V^G_{\bullet,\bullet}\big)\right\},

where $\delta_{Q}(G,V^G_{\bullet,\bullet})$ is defined in <cit.>.
So, there is $Q\in G$ such that $\delta_{Q}(G,V^G_{\bullet,\bullet})<\frac{40}{31}$.

Let $\ell$ is a general line in $G\cong\mathbb{P}^2$ that contains $Q$.
Then  <cit.> gives

\delta_{Q}\big(G,V^G_{\bullet,\bullet}\big)
\geqslant\min\left\{\frac{1}{S\big(V^F_{\bullet,\bullet};\ell\big)},\frac{1}{S\big(W^{G,\ell}_{\bullet,\bullet,\bullet};Q\big)}\right\}.

Let us compute $S(V^G_{\bullet,\bullet};\ell)$ and $S(W^{G,\ell}_{\bullet,\bullet,\bullet};Q)$.
Take $v\in\mathbb{R}_{\geqslant 0}$. Then

&(u-v)\ell\ \text{if $0\leqslant u\leqslant 1$}, \\
&(1-v)\ell\ \text{if $1\leqslant u\leqslant 4$}.
\endaligned
\right.

Let $\mathscr{C}=\widehat{A}\vert_{G}$. Then $\mathscr{C}$ is a smooth cubic curve in $G\cong\mathbb{P}^2$. Let

&0\ \text{if $0\leqslant u\leqslant 1$}, \\
&\frac{u-1}{3}\mathscr{C}\ \text{if $1\leqslant u\leqslant 4$}.
\endaligned
\right.

Now, using <cit.>, we get

\begin{multline*}
S\big(W^{G}_{\bullet,\bullet};\ell\big)=\frac{3}{10}\int_0^4\int_0^\infty \mathrm{vol}\big(P(u)\big\vert_{G}-v\ell\big)dvdu=\\
\end{multline*}

Similarly, it follows from <cit.> that $S(W_{\bullet,\bullet,\bullet}^{G,\ell};Q)$ can be computes as follows:

\begin{multline*}
\frac{3}{10}\int_{0}^{1}\int_{0}^{u}\Big(\big(P(u)\big\vert_{G}-v\ell\big)\cdot\ell\Big)^2dvdu+\frac{3}{10}\int_{1}^{4}\int_{0}^{1}\Big(\big(P(u)\big\vert_{G}-v\ell\big)\cdot\ell\Big)^2dvdu+F_Q=\\
\end{multline*}

where $F_Q=0$ if $Q\not\in\widehat{A}\vert_{G}$, and

F_Q=\frac{6}{10} \int_{1}^{4} \int_{0}^{1}\frac{(1-v)(u-1)}{3}dvdu=\frac{9}{20}

otherwise. This gives $S(W_{\bullet,\bullet,\bullet}^{G,\ell};Q)\leqslant\frac{31}{40}$.
Combining the estimates, we get , which is a contradiction.
This completes the proof of the lemma.

Hence, we see that the surface $A$ has Du Val singularities.
Let $S$ be a general surface in the linear system $|H|$ that contains $P$. Then $S$ is smooth, and $-K_X-uS\sim_{\mathbb{R}} (4-u)H-E$.
Thus, the divisor $-K_X-uS$ is pseudoeffective $\iff$ it is nef $\iff$ $u\leqslant 1$. Then

Let $C=A\vert_{S}$.
Then $C$ is a reduced curve in $|-K_S|$ that is singular at $P$, and $C\cong\pi(C)$.
Moreover, the curve $\pi(C)$ is a general hyperplane section of the cubic surface $\pi(A)\subset\mathbb{P}^3$ that passes through the point $\pi(P)$.
Therefore, since $\pi(A)$ is not a cone by Lemma <ref>, we conclude that the curve $C$ is irreducible.
Hence, one of the following two cases holds:

* the curve $C$ has an ordinary node at $P$,

* the curve $C$ has an ordinary cusp at $P$.

Let $\Pi=\pi(S)$.
Then $\Pi$ is a plane in $\mathbb{P}^3$ such that $\pi(P)\in\Pi$ and $\Pi\cap\pi(A)=\pi(C)$,
and the morphism $\pi\vert_{S}\colon S\to\Pi$ is a composition of blow ups of $9$ intersection points $\Pi\cap\mathscr{C}$,
which we denote by $O_1,\ldots,O_9$.
Note that $\pi(C)$ is a reduced plane cubic curve that passes through these nine points,
and $\pi(C)$ is smooth away from $\pi(P)$.

The curve $C$ cannot have an ordinary double point at the point $P$.

For each $i\in\{1,\ldots,9\}$, let $L_i$ be the proper transform on $S$ of the line in $\Pi$ that passes through  $P$ and $O_i$.
Then $L_i\ne L_j$ for $i\ne j$, since $\pi(C)$ is irreducible. We have

Let $\sigma\colon\widehat{S}\to S$ be the blow up of $S$ at the point $P$,
let $\mathbf{f}$ be the $\sigma$-exceptional curve,
let $\widehat{C},\widehat{L}_1,\ldots,\widehat{L}_{9}$ be  the proper transforms on $\widehat{S}$ of the curves $C,L_1,\ldots,L_9$, respectively.
Then $\widehat{L}_1,\ldots,\widehat{L}_{9}$ are disjoint.
On the surface $\widehat{S}$, we have $\mathbf{f}^2=-1$ and

\widehat{C}^2=-4, \widehat{C}\cdot \widehat{L}_1=\cdots=\widehat{C}\cdot \widehat{L}_9=0, \widehat{C}\cdot\mathbf{f}=2,
\widehat{L}_1^2=\cdots=\widehat{L}_9^2=-1, \widehat{L}_1\cdot\mathbf{f}=\cdots=\widehat{L}_9\cdot\mathbf{f}=1.

Fix $u\in[0,1]$. Let $v$ be a non-negative real number. Then

\begin{equation}
\label{equation:2-4-blow-up-nodal}
\sigma^*\big((-K_X-uS)\big\vert_S\big)-v\mathbf{f}\sim_{\mathbb{R}}\frac{5+u}{6}\widehat{C}+\frac{1-u}{6}\sum_i^9\widehat{L}_i+\frac{19-7u-6v}{6}\mathbf{f}.
\end{equation}

Thus, the divisor $\sigma^*((-K_X-uS)|_S)-v\mathbf{f}$ is pseudoeffective $\iff$ it is nef $\iff$ $v\leqslant\frac{19-7u}{6}$.
Let $P(u,v)$ and $N(u,v)$ be the positive and the negative parts of its Zariski decomposition.
Then, using (<ref>), we compute

&\frac{5+u}{6}\widehat{C}+\frac{1-u}{6}\sum_i^9\widehat{L}_i+\frac{19-7u-6v}{6}\mathbf{f}\ \text{if $0\leqslant v\leqslant \frac{3-3u}{2}$}, \\
&\frac{19-7u-6v}{12}\widehat{C}+\frac{1-u}{6}\sum_i^9\widehat{L}_i+\frac{19-7u-6v}{6}\mathbf{f}\ \text{if $\frac{3-3u}{2}\leqslant v\leqslant 3-u$}, \\
&\frac{19-7u-6v}{12}\big(\widehat{C}+2\sum_i^9\widehat{L}_i+2\mathbf{f}\big)\ \text{if $3-u\leqslant v\leqslant \frac{19-7u}{6}$},
\endaligned
\right.

&0\ \text{if $0\leqslant v\leqslant \frac{3-3u}{2}$}, \\
&\frac{-3+3u+2v}{4}\widehat{C}\ \text{if $\frac{3-3u}{2}\leqslant v\leqslant 3-u$}, \\
&\frac{2v+3u-3}{4}\widehat{C}+(v+u-3)\sum_i^9\widehat{L}_i\ \text{if $3-u\leqslant v\leqslant \frac{19-7u}{6}$},
\endaligned
\right.

\mathrm{vol}\Big(\sigma^*\big((-K_X-uS)\big|_S\big)-v\mathbf{f}\Big)=\left\{\aligned
&(1-u)(7-u)-v^2\ \text{if $0\leqslant v\leqslant \frac{3-3u}{2}$}, \\
&\frac{(1-u)(37-13u-12v)}{4}\ \text{if $\frac{3-3u}{2}\leqslant v\leqslant 3-u$}, \\
&\frac{(19-7u-6v)^2}{4}\ \text{if $3-u\leqslant v\leqslant \frac{19-7u}{6}$},
\endaligned
\right.

&v \ \text{if $0\leqslant v\leqslant \frac{3-3u}{2}$}, \\
&\frac{3(1-u)}{2} \ \text{if $\frac{3-3u}{2}\leqslant v\leqslant 3-u$}, \\
&\frac{3(19-7u-6v)}{2}\ \text{if $3-u\leqslant v\leqslant \frac{19-7u}{6}$}.
\endaligned
\right.

Now, using <cit.>, we get

\begin{multline*}
\frac{3}{10}\int_0^1\int_{0}^{\frac{19-7u}{6}}\mathrm{vol}\Big(\sigma^*\big((-K_X-uS)\big|_S\big)-v\mathbf{f}\Big)dudv=\\
\end{multline*}

Moreover, if $Q$ is a point in $\mathbf{f}$, then <cit.> gives

\begin{multline*}
\end{multline*}

which implies the following assertions:

* if $Q\not\in\widehat{C}\cup\widehat{L}_1\cup\cdots\cup\widehat{L}_{9}$, then $F_Q=0$;

* if $Q\in\widehat{L}_1\cup\cdots\cup\widehat{L}_{9}$, then

* if $Q\in\widehat{C}$ and $\widehat{C}$ intersects $\mathbf{f}$ transversally at $P$, then

\begin{multline*}
\quad \quad \quad \quad F_Q=\frac{6}{10}\int_0^1\int_{\frac{3-3u}{2}}^{3-u}\frac{3(1-u)(2v+3u-3)}{8}dvdu+\\
\frac{6}{10}\int_0^1\int_{3-u}^{\frac{19-7u}{6}}\frac{3(19-7u-6v)(2v+3u-3)}{8}dvdu=\frac{643}{1920};\quad \quad \quad \quad \quad
\end{multline*}

* if $Q\in\widehat{C}$ and $\widehat{C}$ is tangent to $\mathbf{f}$ at the point $P$, then $F_Q=\frac{643}{960}$.

Thus, if $C$ has a node at $P$, then $S(W_{\bullet,\bullet,\bullet}^{\widehat{S},\mathbf{f}};Q)\leqslant\frac{305}{384}$,
so <cit.> gives

1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\min\left\{\inf_{Q\in\mathbf{f}}\frac{1}{S(W_{\bullet,\bullet,\bullet}^{\widehat{S},\mathbf{f}};Q)},
\frac{2}{S(V_{\bullet,\bullet}^{S};\mathbf{f})},\frac{1}{S_X(S)}\right\}
\geqslant\min\left\{\frac{384}{305},\frac{960}{767},\frac{40}{13}\right\}=\frac{960}{767}>1,

which is a contradiction. This shows that $C$ has a cusp at $P$.

The curve $C$ cannot have an ordinary cusp at the point $P$.

Suppose $C$ has a cusp. Let $L$ be an irreducible curve in $S$ such that $\pi(L)$ is a line and $\pi(L)\cap\pi(C)=\pi(P)$.
Then $(-K_X-uS)\vert_{S}\sim_{\mathbb{R}}(1-u)L+C$.

Now, we consider the following commutative diagram:

\xymatrix{
S &&&& \widehat{S}\ar@{->}[llll]_{\sigma}}

where $\sigma_1$ is the blow up of $P$,
$\sigma_2$ is the blow up of the point in the $\sigma_1$-exceptional curve contained in the proper transform of $C$,
$\sigma_3$ is the blow up of the point in the $\sigma_2$-exceptional curve contained in the proper transform of $C$,
$\upsilon$ is the birational contraction of the proper transforms of $\sigma_1\circ\sigma_2$-exceptional curves,
and $\sigma$ is the birational contraction of the proper transform of the $\sigma_3$-exceptional curve.
Then $\widehat{S}$ has two singular points:

* a cyclic quotient singularity of type $\frac{1}{2}(1,1)$, which we denote by $Q_2$;

* a cyclic quotient singularity of type $\frac{1}{3}(1,1)$, which we denote by $Q_3$.

Let $\mathbf{f}$ be the $\sigma$-exceptional curve,
let $\widehat{C}$ be the proper transform on $\widehat{S}$ of the curve $C$,
and let $\widehat{L}$ be the proper transform of the curve $L$.
Then  the curves $\mathbf{f}$, $\widehat{C}$, $\widehat{L}$ are smooth.
Moreover, it is not very difficult to check that $Q_2\in\mathbf{f}\ni Q_3$, $Q_2\not\in\widehat{C}\not\ni Q_3$, $Q_2\in\widehat{L}\not\ni Q_2$.
Further, we have $A_S(\mathbf{f})=5$, $\sigma^*(C)\sim\widehat{C}+6\mathbf{f}$, $\sigma^*(L)\sim\widehat{L}+3\mathbf{f}$.
On the surface $\widehat{S}$, we have

\widehat{L}^2=-\frac{1}{2}, \widehat{L}\cdot\widehat{C}=0, \widehat{C}\cdot \mathbf{f}=\frac{1}{2}, \widehat{C}^2=-6, \widehat{C}\cdot \mathbf{f}=1, \mathbf{f}^2=-\frac{1}{6}.

Note that  $Q_2=\mathbf{f}\cap\widehat{L}$, and $\widehat{C}$ intersects $\mathbf{f}$ transversally by one point.

Fix $u\in[0,1]$. Let $v$ be a non-negative real number. Then

\begin{equation}
\label{equation:2-4-blow-up-cuspidal}
\sigma^*\big((-K_X-uS)\big\vert_{S}\big)-v\mathbf{f}\sim_{\mathbb{R}}(1-u)\widehat{L}+\widehat{C}+(9-3u-v)\mathbf{f}.
\end{equation}

Thus, the divisor $\sigma^*((-K_X-uS)|_S)-v\mathbf{f}$ is pseudoeffective $\iff$ it is nef $\iff$ $v\leqslant 9-3u$.
Let $P(u,v)$ and $N(u,v)$ be the positive and the negative parts of its Zariski decomposition.
Then using (<ref>), we compute

&(1-u)\widehat{L}+\widehat{C}+(9-3u-v)\mathbf{f}\ \text{if $0\leqslant v\leqslant 3-3u$}, \\
&(1-u)\widehat{L}+\frac{9-3u-v}{6}\widehat{C}+(9-3u-v)\mathbf{f}\ \text{if $3-3u\leqslant v\leqslant 8-2u$}, \\
&\frac{9-3u-v}{6}\big(6\widehat{L}+\widehat{C}+6\mathbf{f}\big)\ \text{if $8-2u\leqslant v\leqslant 9-3u$},
\endaligned
\right.

&0\ \text{if $0\leqslant v\leqslant 3-3u$}, \\
&\frac{v+3u-3}{6}\widehat{C}\ \text{if $3-3u\leqslant v\leqslant 8-2u$}, \\
&\frac{v+3u-3}{6}\widehat{C}+(v+2u-8)\widehat{L}\ \text{if $8-2u\leqslant v\leqslant 9-3u$}.
\endaligned
\right.

This gives

\mathrm{vol}\Big(\sigma^*\big((-K_X-uS)\big|_S\big)-v\mathbf{f}\Big)=\left\{\aligned
&(1-u)(7-u)-\frac{v^2}{6}\ \text{if $0\leqslant v\leqslant 3-3u$}, \\
&\frac{(1-u)(17-5u-2v)}{2}\ \text{if $3-3u\leqslant v\leqslant 8-2u$}, \\
&\frac{(9-3u-v)^2}{2}\ \text{if $8-2u\leqslant v\leqslant 9-3u$},
\endaligned
\right.

&\frac{v}{6}\ \text{if $0\leqslant v\leqslant 3-3u$}, \\
&\frac{1-u}{2}\ \text{if $3-3u\leqslant v\leqslant 8-2u$}, \\
&\frac{9-3u-v}{2}\ \text{if $8-2u\leqslant v\leqslant 9-3u$}.
\endaligned
\right.

Now, we use <cit.> to get

\begin{multline*}
\end{multline*}

Similarly, if $Q$ is a point in $\mathbf{f}$, then <cit.> gives

\begin{multline*}
\end{multline*}

where $F_Q$ can be computes as follows:

* if $Q\ne\widehat{C}\cap\mathbf{f}$ and $Q\ne\widehat{L}\cap\mathbf{f}$, then $F_Q=0$;

* if $Q=\widehat{L}\cap\mathbf{f}$, then

* if $Q=\widehat{C}\cap\mathbf{f}$ , then

\begin{multline*}
\quad \quad \quad \quad
+\frac{6}{10}\int_0^1\int_{8-2u}^{9-3u}\frac{(9-3u-v)(v+3u-3)}{12}dudv=\frac{193}{480}.\quad \quad \quad \quad\quad
\end{multline*}

Therefore, we conclude that

\left\{\aligned
&\frac{5}{32}\ \text{if $Q\not\in\widehat{C}\cup\widehat{L}$}, \\
&\frac{27}{80} \ \text{if $Q=\widehat{L}\cap\mathbf{f}$}, \\
&\frac{67}{120}\ \text{if $Q=\widehat{C}\cap\mathbf{f}$}.
\endaligned
\right.

Let $\Delta_\mathbf{f}=\frac{1}{2}Q_2+\frac{2}{3}Q_3$. Then, using <cit.> and our computations, we get

1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\min\left\{\inf_{Q\in\mathbf{f}}\frac{1-\mathrm{ord}_Q\big(\Delta_\mathbf{f}\big)}{S(W_{\bullet,\bullet,\bullet}^{\widehat{S},\mathbf{f}};Q)},
\frac{5}{S(V_{\bullet,\bullet}^{S};\mathbf{f})},\frac{1}{S_X(S)}\right\}
\geqslant \min \left\{\frac{40}{13}, \frac{200}{173}, \frac{120}{67}\right\}=\frac{200}{173}>1,

which is a contradiction.

Combining Lemmas <ref> and <ref>, we obtain a contradiction.

All smooth Fano threefolds in the family №2.4 are K-stable.

§ FAMILY №2.6 (VERRA THREEFOLDS)

Smooth Fano threefolds in the family №2.6 can be described as follows:

(a) smooth divisors of degree $(2,2)$ in $\mathbb{P}^2\times \mathbb{P}^2$, which are known as Verra threefolds,

(b) double covers of the (unique) smooth divisor in $\mathbb{P}^2\times \mathbb{P}^2$ of degree $(1,1)$ branched over smooth anticanonical K3 surfaces.

Note that every double cover of the smooth divisor in $\mathbb{P}^2\times \mathbb{P}^2$ of degree $(1,1)$ branched over a smooth anticanonical surface is K-stable <cit.>.
In fact, this also implies that general Verra threefold is K-stable <cit.>.

The goal of this section is to prove that all smooth Verra threefolds are K-stable.

Let $X$ be a smooth divisor of degree $(2,2)$ in $\mathbb{P}^2\times \mathbb{P}^2$,
let $\pi_1\colon X\to \mathbb{P}^2$ be the projection to the first factor of $\mathbb{P}^2\times \mathbb{P}^2$,
and let $\pi_2\colon X\to \mathbb{P}^2$ be the projection to the second factor.
Then $\pi_1$ and $\pi_2$ are conic bundles [18].
Set $H_1=\pi_1^*(\mathcal{O}_{\mathbb{P}^2}(1))$ and $H_2=\pi_2^*(\mathcal{O}_{\mathbb{P}^2}(1))$.

-K_X\sim H_1+H_2,

and the group $\mathrm{Pic}(X)$ is generated by $H_1$ and $H_2$.
Note that the group $\mathrm{Aut}(X)$ is finite [9].
Thus, the threefold $X$ is K-stable $\iff$ it is K-polystable. See [19] for details.

Fix a point $P\in X$. Let $C_1$ be the fiber of the conic bundle $\pi_1$ that contains $P$,
and let $C_2$ be the fiber of the conic bundle $\pi_2$ contains $P$.
Then $C_1$ or $C_2$ is smooth at $P$.

Local computations.

Let $\Delta_1$ and $\Delta_2$ be the discriminant curves of the conic bundles $\pi_1$ and $\pi_2$, respectively.
Then  $\Delta_1$ and $\Delta_2$ are reduced curves of degree $6$
that have at most ordinary double points as singularities.
For basic properties of the discriminant curves  $\Delta_1$ and $\Delta_2$, see .
In particular, we know that no line in $\mathbb{P}^2$ can be an irreducible component of these curves.

Fix a point $P\in X$. Let $C_2$ be the fiber of the conic bundle $\pi_2$ that contains $P$,
let $S$ be a general surface in $|H_2|$ that contains $C_2$.
Then one of the following cases holds:

* $C_2$ is smooth, $S$ is smooth, the divisor $-K_S$ is ample;

* $C_2$ is smooth, $S$ is smooth, the divisor $-K_S$ is nef, the surface $S$ has exactly four irreducible curves that
have trivial intersection with the divisor $-K_S$, these curves are disjoint and none of them passes through $P$,
and $C_2\subset\mathrm{Sing}(\pi_1^{-1}(\Delta_1))$;

* $C_2$ is singular and reduced, $S$ is smooth, the divisor $-K_S$ is ample;

* $C_2$ is not reduced, $P\not\in\mathrm{Sing}(S)\subset\mathrm{Supp}(C_2)$,
and $\mathrm{Sing}(S)$ consists of two points, which are ordinary double points of the surface $S$,
and $-K_S$ is ample.

The assertion about the singularities of the surface $S$ are local and well-known.

We have $-K_S\sim H_1\vert_S$ and $K_S^2=2$, so  $S$ is a weak del Pezzo surface of degree $2$,
and the restriction $\pi_1\vert_{S}\colon S\to\mathbb{P}^2$ is the anticanonical morphism.
Let $\ell$ be an irreducible curve in the surface $S$ such that

-K_S\cdot \ell=0.

Then $\ell$ is an irreducible component of the fiber $\pi_1^{-1}(\pi_1(\ell))$.
But $\pi_2(\ell)$ is the line $\pi_2(S)$, which implies that the scheme fiber $\pi_1^{-1}(\pi_1(\ell))$ is singular,
which implies that $\pi_1(\ell)\in\Delta_1$.
Since $\ell_1\cap C_2\ne\varnothing$ and $\pi_2(S)$ is a general line in $\mathbb{P}^2$ that passes through the point $\pi_2(P)$,
we see that $\pi_1(C_2)\subset\Delta_1$. This implies that $C_2$ is irreducible and reduced.

Let $R=\pi_1^{-1}(\pi_1(C_2))$. Then the surface $R$ is singular along a curve that is isomorphic to the conic $\pi_1(C_2)\cong C_2$.
Let $f\colon\widetilde{R}\to R$ be the blow up of this curve. Then $\widetilde{R}$ is smooth,
and the composition morphism $\pi_1\circ f$ induces a $\mathbb{P}^1$-bundle $\widetilde{R}\to\widetilde{C}_2$,
where $\widetilde{C}_2$ is double cover of the conic $\pi_1(C_2)$ that is branched over the eight points $\pi_1(C_2)\cap(\Delta_1-\pi_1(C_2))$.
In particular, we see that $\widetilde{C}_2$ is an irrational curve, which implies that

Vice versa, if the fiber $C_2$ is a smooth conic, and the conic $\pi_1(C_2)$ is an irreducible component of the discriminant curve $\Delta_1$,
then it follows from the Bertini theorem that

where $\ell_1,\ell_2,\ldots,\ell_k$ are $k$ distinct irreducible reduced curves in $X$ that are irreducible components of the fibers
of the natural projection  $\pi_1^{-1}(\pi_1(C_2))\to \pi_1(C_2)$.

4=2H_2^2\cdot H_1=H_2\cdot S\cdot\pi_1^{-1}(\pi_1(C_2))=H_2\cdot\Big(2C_2+\sum_{i=1}^k\ell_i\Big)=k,

we see that $S$ contains exactly $4$ irreducible curves that intersects trivially with $-K_S$.
Now, the generality in the choice of $S$ implies that none of these curves contains $P$.

Actually, the case (2) in Lemma <ref> can happen.
Indeed, in the assumption and notations of Lemma <ref>, let $P=([0:0:1],[0:0:1])$,
and suppose that $X$ is given by

where $([u:v:w],[x:y:z])$ are coordinates on $\mathbb{P}^2\times\mathbb{P}^2$.
Then the threefold $X$ is smooth.
For instance, this can be checked using the following Magma script:

Observe that the fiber $C_1$ is a singular reduced curve given by $u=v=2x^2-xy+y^2=0$,
the fiber $C_2$ is a smooth curve that is given by $x=y=uw+v^2=0$,
and the discriminant curve $\Delta_1$ is a union of the conic $\pi_1(C_2)$ and the irreducible quartic plane curve given by

\begin{multline*}
\end{multline*}

As in case (2) in Lemma <ref>, we have $C_2=\mathrm{Sing}(\pi_1^{-1}(\pi_1(C_2)))$.

Let us prove that $X$ is K-stable. Suppose it is not.
Using the valuative criterion [14, 16],
we see that there exists a prime divisor $\mathbf{F}$ over $X$ such that

\beta(\mathbf{F})=A_X(\mathbf{F})-S_X(\mathbf{F})\leqslant 0.

Let $Z$ be the center of the divisor $\mathbf{F}$ on $X$.
Then $Z$ is not a surface by <cit.>.

Let $P$ be any point in $Z$,
let $C_1$ be the fiber of the conic bundle $\pi_1$ that contains $P$,
and let $C_2$ be the fiber of the conic bundle $\pi_2$ that contains $P$.
By Lemma <ref>, at least one curve among $C_1$ or $C_2$ is smooth at $P$.
We may assume that $C_2$ is smooth at $P$.

Let $S$ be a general surface in $|H_2|$ that contains $C_2$.
Then $S$ is smooth by Lemma <ref>.
Moreover, one of the following three cases holds:

* $C_2$ is smooth, the divisor $-K_S$ is ample;

* $C_2$ is smooth, $\pi_1(C_2)\subset\Delta_1$, the divisor $-K_S$ is nef, the surface $S$ has exactly four irreducible curves that
have trivial intersection with the divisor $-K_S$, these curves are disjoint and none of them passes through $P$;

* $C_2$ is singular and reduced, the divisor $-K_S$ is ample.

Let $C$ be the curve in $X$ that is defined as follows:

* if $C_2$ is smooth and irreducible, we let $C=C_2$.

* if $C_2$ is reducible, we let $C$ be its irreducible component that contains $P$.

Note that $-K_S\sim H_1\vert_{S}$ and $K_S^2=2$.
Let $\eta\colon S\to\mathbb{P}^2$ be the restriction morphism $\pi_1\vert_{S}$.
Then $\eta$ is an anticanonical morphism of the surface $S$, which is generically two-to-one.
Hence, the morphism $\eta$ induces an involution $\tau\in\mathrm{Aut}(S)$. We let $C^\prime=\tau(C)$.

Now, let $u$ be a non-negative real number. Then we have $-K_X-uS\sim_{\mathbb{R}} H_1+(1-u)H_2$,
so the divisor $-K_X-uS$ is nef $\iff$ it is pseudoeffective $\iff$ $u\leqslant 1$. Then $S_X(S)=\frac{5}{12}$.
Now, let us use notations introduced in <cit.>.
Using <cit.>, we get

1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant \min\left\{\frac{1}{S_X(S)},\frac{1}{S(W_{\bullet,\bullet}^S;C)},\frac{1}{S(W_{\bullet, \bullet,\bullet}^{S,C};P)}\right\},

where $S(W_{\bullet,\bullet}^S;C)$ and $S(W_{\bullet, \bullet,\bullet}^{S,C};P)$ are defined in <cit.>.
Hence, since $S_X(S)<1$, we get

\begin{equation}
\label{equation:2-6-main-inequality}
\max\left\{S(W_{\bullet,\bullet}^S;C),S(W_{\bullet, \bullet,\bullet}^{S,C};P)\right\}\geqslant 1.
\end{equation}

Moreover, if $Z=P$, then it also follows from <cit.> that

\begin{equation}
\label{equation:2-6-main-inequality-refined}
\max\left\{S(W_{\bullet,\bullet}^S;C),S(W_{\bullet, \bullet,\bullet}^{S,C};P)\right\}>1.
\end{equation}

Let us estimate  $S(W_{\bullet,\bullet}^S;C)$ and $S(W_{\bullet, \bullet,\bullet}^{S,C};P)$ using results obtained in <cit.>.

Let $P(u)=-K_X-uS$. Then <cit.> gives

S\big(W^S_{\bullet,\bullet};C\big)=\frac{1}{4}\int_0^1\int_0^\infty \mathrm{vol}\big(P(u)\big\vert_{S}-vC\big)dvdu,

Let $P(u,v)$ be the positive part of the Zariski decomposition of the divisor $P(u)\vert_{S}-vC$,
and let $N(u,v)$ be its negative part,  where $u\in[0,1]$ and $v\in\mathbb{R}_{\geqslant 0}$.

S\big(W_{\bullet,\bullet,\bullet}^{S,C};P\big)=F_{P}+\frac{1}{4}\int_{0}^{1}\int_{0}^{\infty}\big(P(u,v)\cdot C\big)^2dvdu

by <cit.>, where

F_P=\frac{1}{2}\int_{0}^{1}\int_{0}^{\infty}\big(P(u,v)\cdot C\big)\mathrm{ord}_P\Big(N(u,v)\big\vert_{C}\Big)dvdu.

Suppose that $C_2$ is smooth, and $-K_S$ is ample. Then

\begin{align*}
S(W_{\bullet, \bullet,\bullet}^{S,C};P)&=1.
\end{align*}

We have $C\cdot C^\prime=4$ and $(C^\prime)^2=C^2=0$. Fix $u\in[0,1]$ and $v\in\mathbb{R}_{\geqslant 0}$. Then

which implies that $P(u)\vert_{S}-vC$ is pseudoeffective $\iff$ $P(u)\vert_{S}-vC$ is nef $\iff$ $v\leqslant \frac{3}{2}-u$.
If $0\leqslant u\leqslant 1$ and $0\leqslant v\leqslant \frac{3}{2}-u$, then $P(u,v)=(\frac{3}{2}-u-v)C+\frac{1}{2}C^\prime$ and $N(u,v)=0$, so

\frac{1}{4}\int_0^1\int_0^{\frac{3}{2}-u}6-4u-4v\,dvdu=\frac{13}{24}.

Similarly, we see that $F_P=0$ and $P(u,v)\cdot C=2$, which gives $S(W_{\bullet,\bullet,\bullet}^{S,C};P)=1$.

Suppose that $C_2$ is smooth, $-K_S$ is not ample. Then

\begin{align*}
S(W_{\bullet, \bullet,\bullet}^{S,C};P)&=\frac{5}{6}.
\end{align*}

In this case, we have the following commutative diagram:

\xymatrix{
\overline{S}\ar@{->}[rr]^{\pi}&&\mathbb{P}^2}

where $\phi$ is a birational map that contracts four disjoint $(-2)$-curves,
and $\overline{S}$ is a del Pezzo surface of degree $2$ that has $4$ isolated ordinary double points,
and $\pi$ is a double cover that is ramified in a reducible quartic curve that is a union of two irreducible conics
such that $\eta(C)$ is one of these two conics.
In particular, we have $C=\tau(C)$.

Let $E_1$, $E_2$, $E_3$, $E_4$ be the $\phi$-exceptional curves.
Fix $u\in[0,1]$ and $v\in\mathbb{R}_{\geqslant 0}$. Then

so the divisor $P(u)\vert_{S}-vC$ is pseudoeffective $\iff$ $v\leqslant 2-u$.
Moreover, we have

&(2-u-v)C+\frac{1}{2}(E_1+E_2+E_3+E_4)\ \text{if $0\leqslant v\leqslant 1-u$}, \\
&(2-u-v)\Big(C+\frac{1}{2}(E_1+E_2+E_3+E_4)\Big)\ \text{if $1-u\leqslant v\leqslant 2-u$}, \\
\endaligned
\right.

&0\ \text{if $0\leqslant v\leqslant 1-u$}, \\
&\frac{u+v-1}{2}(E_1+E_2+E_3+E_4)\ \text{if $1-u\leqslant v\leqslant 2-u$}, \\
\endaligned
\right.

\mathrm{vol}\big(P(u)\vert_{S}-vC\big)=
\left\{\aligned
&(6-4u)-4\ \text{if $0\leqslant v\leqslant 1-u$}, \\
&2(2-u-v)^2\ \text{if $1-u\leqslant v\leqslant 2-u$}. \\
\endaligned
\right.

Now, integrating $\mathrm{vol}(P(u)\vert_{S}-vC)$, we obtain $S(W_{\bullet,\bullet}^S;C)=\frac{7}{12}$.

To compute $S(W_{\bullet,\bullet,\bullet}^{S,C};P)$, we first observe that $F_{P}=0$,
because $P\not\in E_1\cup E_2\cup E_3\cup E_4$, since $S$ is a general surface in $|H_2|$ that contains  $C_2$.
On the other hand, we have

P(u,v)\cdot C=\left\{\aligned
&2\ \text{if $0\leqslant v\leqslant 1-u$}, \\
&4-2u-2v\ \text{if $1-u\leqslant v\leqslant 2-u$}. \\
\endaligned
\right.

Hence, integrating $(P(u,v)\cdot C)^2$, we get $S(W_{\bullet,\bullet,\bullet}^{S,C};P)=\frac{5}{6}$ as required.

Suppose that $C_2$ is singular. Then

\begin{align*}
S(W_{\bullet, \bullet,\bullet}^{S,C};P)&\leqslant \frac{11}{12}.
\end{align*}

The curve $C_2$ consists of two irreducible components: the curve $C$ and another curve, which we denote by $L$.
The curves $C$ and $L$ are smooth and intersects transversally at one point.
Note that $P\ne C\cap L$, since $C_2$ is smooth at the point $P$ by assumption.

The intersections of the curves $C$, $L$ and $C^\prime=\tau(C)$ on $S$ are given in the table below.

$\bullet$      $C$     $L$     $C^\prime$

$C$            $-1$     $1$     $2$

$L$            $1$     $-1$     $0$

$C^\prime$     $2$     $0$     $-1$

Fix $u\in[0,1]$ and $v\in\mathbb{R}_{\geqslant 0}$. Since $C+C^\prime\sim -K_S$, we have

so $P(u)\vert_{S}-vC$ is pseudoeffective $\iff$ $v\leqslant 2-u$.
Moreover, if $0\leqslant u\leqslant\frac{1}{2}$, then

&(2-u-v)C+(1-u)L+C^\prime\ \text{if $0\leqslant v\leqslant 1$}, \\
&(2-u-v)(C+L)+C^\prime\ \text{if $1\leqslant v\leqslant \frac{3}{2}-u$}, \\
&(2-u-v)(C+L+C^\prime)\ \text{if $\frac{3}{2}-u\leqslant v\leqslant 2-u$},
\endaligned
\right.

&0\ \text{if $0\leqslant v\leqslant 1$}, \\
&(v-1)L\ \text{if $1\leqslant v\leqslant \frac{3}{2}-u$}, \\
&(v-1)L+(2v+2u-3)C^\prime\ \text{if $\frac{3}{2}-u\leqslant v\leqslant 2-u$},
\endaligned
\right.

\mathrm{vol}\big(P(u)\vert_{S}-vC\big)=
\left\{\aligned
&6-v^2-4u-2v\ \text{if $0\leqslant v\leqslant 1$}, \\
&7-4u-4v \ \text{if $1\leqslant v\leqslant \frac{3}{2}-u$}, \\
&4(u+v-2)^2\ \text{if $\frac{3}{2}-u\leqslant v\leqslant 2-u$}.
\endaligned
\right.

Similarly, if $\frac{1}{2}\leqslant u\leqslant 1$, then

&(2-u-v)C+(1-u)L+C^\prime\ \text{if $0\leqslant v\leqslant \frac{3}{2}-u$}, \\
&(2-u-v)(C+2C^\prime)+(1-u)L\ \text{if $\frac{3}{2}-u\leqslant v\leqslant 1$}, \\
&(2-u-v)(C+L+C^\prime)\ \text{if $1\leqslant v\leqslant 2-u$},
\endaligned
\right.

&0\ \text{if $0\leqslant v\leqslant \frac{3}{2}-u$}, \\
&(2v+2u-3)C^\prime\ \text{if $\frac{3}{2}-u\leqslant v\leqslant 1$}, \\
&(v-1)L+(2v+2u-3)C^\prime\ \text{if $1\leqslant v\leqslant 2-u$},
\endaligned
\right.

\mathrm{vol}\big(P(u)\vert_{S}-vC\big)=
\left\{\aligned
&6-v^2-4u-2v\ \text{if $0\leqslant v\leqslant \frac{3}{2}-u$}, \\
&(5-2u-3v)(3-2u-v)\ \text{if $\frac{3}{2}-u\leqslant v\leqslant 1$}, \\
&4(u+v-2)^2\ \text{if $1\leqslant v\leqslant 2-u$}.
\endaligned
\right.

Hence, integrating $\mathrm{vol}(P(u)\vert_{S}-vC)$, we get $S(W_{\bullet,\bullet}^S;C)=\frac{3}{4}$.

Now, let us compute $S(W_{\bullet, \bullet,\bullet}^{S,C};P)$.
If $0\leqslant u\leqslant \frac{1}{2}$, then

P(u,v)\cdot C=\left\{\aligned
&1+v\ \text{if $0\leqslant v\leqslant 1$}, \\
&2\ \text{if $1\leqslant v\leqslant \frac{3}{2}-u$}, \\
&8-4u-4v\ \text{if $\frac{3}{2}-u\leqslant v\leqslant 2-u$}.
\endaligned
\right.

Similarly, if $\frac{1}{2}\leqslant u\leqslant 1$, then

P(u,v)\cdot C=\left\{\aligned
&1+v\ \text{if $0\leqslant v\leqslant \frac{3}{2}-u$}, \\
&7-4u-3v\ \text{if $\frac{3}{2}-u\leqslant v\leqslant 1$}, \\
&8-4u-4v\ \text{if $1\leqslant v\leqslant 2-u$}.
\endaligned
\right.

Then, integrating, we get $S(W_{\bullet,\bullet,\bullet}^{S,C};P)=\frac{145}{192}+F_{P}$.
To compute $F_{P}$, let us recall that $P\not\in L$.
Hence, if $P\not\in C^\prime$, then $F_{P}=0$,
which implies that $S(W_{\bullet,\bullet,\bullet}^{S,C};P)=\frac{145}{192}<\frac{11}{12}$ as required.

We may assume that $P\in C\cap C^\prime$.
If $C^\prime$ intersects $C$ transversally at $P$, then

\mathrm{ord}_P\Big(N(u,v)\big\vert_{C}\Big)=\left\{\aligned
&0\ \text{if $0\leqslant v\frac{3}{2}-u$}, \\
&2u+2v-3\ \text{if $\frac{3}{2}-u\leqslant v\leqslant 2-u$},
\endaligned
\right.

which gives

\begin{multline*}
\quad \quad \quad \quad F_P=\frac{1}{2}\int_{0}^{\frac{1}{2}}\int_{\frac{3}{2}-u}^{2-u}(8-4u-4v)(2u+2v-3)dvdu+\\
\end{multline*}

so $S(W_{\bullet,\bullet,\bullet}^{S,C};P)=\frac{145}{192}+\frac{31}{384}=\frac{107}{128}<\frac{11}{12}$.
If $C^\prime$ is tangent to $C$ at the point $P$, then

\mathrm{ord}_P\Big(N(u,v)\big\vert_{C}\Big)=\left\{\aligned
&0\ \text{if $0\leqslant v\frac{3}{2}-u$}, \\
&2(2u+2v-3)\ \text{if $\frac{3}{2}-u\leqslant v\leqslant 2-u$},\\
\endaligned
\right.

which gives $F_P=\frac{31}{192}$, so $S(W_{\bullet,\bullet}^{S,C};P)=\frac{145}{192}+\frac{31}{192}=\frac{11}{12}$.

Now, using (<ref>) and Lemmas <ref>, <ref>, <ref>,
we see that $Z$ is a curve.

One has $H_1\cdot Z\geqslant 1$ and $H_2\cdot Z\geqslant 1$.

If $H_2\cdot Z=0$, then $Z=C$, which is impossible by (<ref>) and Lemmas <ref>, <ref>, <ref>.
Hence, we see that $H_2\cdot Z\geqslant 1$ and $\pi_2(Z)$ is a curve.
Let us show that $H_1\cdot Z\geqslant 1$.

Suppose that $H_1\cdot Z=0$. Then $Z$ must be an irreducible component of the curve $C_1$.
If $C_1$ is reduced, then arguing exactly as in the proofs of Lemmas <ref>, <ref>, <ref>,
we obtain a contradiction with <cit.>. Thus, we see that $C_1$ is not reduced.

So far, the point $P$ was a point in $Z$. Let us choose $P\in Z$ such that $\pi_2(P)\in\pi_2(Z)\cap\Delta_2$.
Then $C_2$ is singular. But it is smooth at $P$ by Lemma <ref>,
which fits our assumption above.
Then $S(W_{\bullet,\bullet}^S;C)=\frac{3}{4}$ and $S(W_{\bullet, \bullet,\bullet}^{S,C};P)\leqslant \frac{11}{12}$ by Lemma <ref>, which contradicts (<ref>).

Both $\pi_1(Z)$ and $\pi_2(Z)$ are curves.
Similar to what we did in the proof of Lemma <ref>, let us choose the point $P\in Z$ such that $\pi_1(P)\in\Delta_1$.
Then $C_1$ is singular at $P$, which implies that $C_2$ is smooth at $P$ by Lemma <ref>.
Now, using (<ref>) and Lemmas <ref> and <ref>, we see that
$C=C_2$, the curve $C_2$ is smooth, the divisor $-K_S$ is ample.

We see that $S$ is a del Pezzo surface, and $\eta\colon S\to\mathbb{P}^2$ is a double cover ramified in a smooth quartic curve,
so we are almost in the same position as in the proof of Lemma <ref>.
But now, we have one small advantage: the point $\eta(P)$ is contained in the ramification divisor of the double cover $\eta$,
because $C_1$ is singular at $P$.
Then $|-K_S|$ contains a unique curve that is singular at $P$.
Denote this curve by $R$. We have the following possibilities:

(a) $R$ is an irreducible curve that has a nodal singularity at $P$;

(b) $R$ is an irreducible curve that has a cuspidal singularity at $P$;

(c) $R=R_1+R_2$ for two $(-1)$-curves in $S$ that intersect transversally at $P$;

(d) $R=R_1+R_2$ for two $(-1)$-curves in $S$ that are tangent at $P$.

Let $f\colon\widetilde{S}\to S$ be the blow up of the point $P$.
Denote by $\widetilde{R}$ and $\widetilde{C}$ the proper transforms on the surface $\widetilde{S}$ of the curves $R$ and $C$, respectively.
Fix $u\in[0,1]$ and $v\in\mathbb{R}_{\geqslant 0}$. Then

f^*\big(P(u)\vert_{S}\big)-vE\sim_{\mathbb{Q}} \widetilde{R}+(1-u)\widetilde{C}+(3-u-v)E.

Let $\widetilde{P}(u,v)$ be the positive part of the Zariski decomposition of $\widetilde{R}+(1-u)\widetilde{C}+(3-u-v)E$,
and let $\widetilde{N}(u,v)$ its negative part.
Then it follows from <cit.> that

\begin{equation}
\label{equation:2-6-final}
1\geqslant \frac{A_X(\mathbf{F})}{S_X(\mathbf{F})}\geqslant\min\left\{\frac{1}{S_X(S)},
\frac{2}{S(W_{\bullet,\bullet}^{\widetilde{S}};E)},\inf_{O\in E}\frac{1}{S(W_{\bullet, \bullet,\bullet}^{\widetilde{S},E};O)}\right\},
\end{equation}

where $S(W_{\bullet,\bullet}^{\widetilde{S}};E)$ and $S(W_{\bullet, \bullet,\bullet}^{\widetilde{S},E};O)$
are defined  in [3] similar to $S(W_{\bullet,\bullet}^S;C)$ and $S(W_{\bullet, \bullet,\bullet}^{S,C};P)$.
These two numbers can be computed using   <cit.>. Namely, we have

S\big(W_{\bullet, \bullet,\bullet}^{\widetilde{S},E};O\big)=\frac{1}{4}\int_0^1\int_0^{\infty}\Big(\big(\widetilde{P}(u,v)\cdot E\big)\Big)^2dvdu+

where $O$ is a point in $E$ and

F_O=\frac{1}{2}\int_0^1\int_0^{\infty}\big(\widetilde{P}(u,v)\cdot E\big)\mathrm{ord}_O\Big(\widetilde{N}(u,v)\big|_{E}\Big)dvdu.

Let us use these formulas to estimate $S(W_{\bullet,\bullet}^{\widetilde{S}};E)$ and $S(W_{\bullet, \bullet,\bullet}^{\widetilde{S},E};O)$.

If the curve $R$ is irreducible, the intersections of the curves $\widetilde{C}$, $\widetilde{R}$, $E$ can be computed as follows:
$\widetilde{C}^2=-1$, $\widetilde{C}\cdot\widetilde{R}=0$, $\widetilde{C}\cdot E=1$, $\widetilde{R}^2=-2$, $\widetilde{R}\cdot E=2$, $E^2=-1$.
If $R$ is reducible, then $\widetilde{R}=\widetilde{R}_1+\widetilde{R}_2$ for two smooth irreducible curves $\widetilde{R}_1+\widetilde{R}_2$
such that the intersection form of the curves $\widetilde{C}$, $\widetilde{R}_1$,  $\widetilde{R}_1$ and $E$ is given in the following table:

$\bullet$      $\widetilde{C}$     $\widetilde{R}_1$     $\widetilde{R}_1$     $E$

$\widetilde{C}$            $-1$     $0$     $0$     $1$

$\widetilde{R}_1$            $0$     $-2$     $1$     $1$

$\widetilde{R}_1$            $0$     $1$     $-2$     $1$

$E$     $1$     $1$     $1$     $-1$

Then $\widetilde{R}+(1-u)\widetilde{C}+(3-u-v)E$ is pseudoeffective $\iff$ $v\leqslant 3-u$.
Moreover, we have

\widetilde{P}(u,v)=\left\{\aligned
&\widetilde{R}+(1-u)\widetilde{C}+(3-u-v)E\ \text{if $0\leqslant v\leqslant 2-u$}, \\
&(1-u)\widetilde{C}+(3-u-v)(E+\widetilde{R})\ \text{if $2-u\leqslant v\leqslant 2$}, \\
&(3-u-v)(E+\widetilde{R}+\widetilde{C})\ \text{if $2\leqslant v\leqslant 3-u$}, \\
\endaligned
\right.

\widetilde{N}(u,v)=\left\{\aligned
&0\ \text{if $0\leqslant v\leqslant 2-u$}, \\
&(v+u-2)\widetilde{R}\ \text{if $2-u\leqslant v\leqslant 2$}, \\
&(v+u-2)\widetilde{R}+(v-2)\widetilde{C}\ \text{if $2\leqslant v\leqslant 3-u$},\\
\endaligned
\right.

\big(\widetilde{P}(u,v)\big)^2=
\left\{\aligned
&6-v^2-4u\ \text{if $0\leqslant v\leqslant 2-u$}, \\
&14+2u^2+4uv+v^2-12u-8v\ \text{if $2-u\leqslant v\leqslant 2$}, \\
&2(3-u-v)^2\ \text{if $2\leqslant v\leqslant 3-u$}, \\
\endaligned
\right.

Now, integrating, we obtain $S(W_{\bullet,\bullet}^{\widetilde{S}};E)=\frac{17}{12}$.

Fix a point $O\in E$. To estimate $S(W_{\bullet, \bullet,\bullet}^{\widetilde{S},E};O)$, first we observe that

\widetilde{P}(u,v)\cdot E=
\left\{\aligned
&v\ \text{if $0\leqslant v\leqslant 2-u$}, \\
&4-2u-v\ \text{if $2-u\leqslant v\leqslant 2$}, \\
&6-2u-2v\ \text{if $2\leqslant v\leqslant 3-u$}.\\
\endaligned
\right.

Therefore, integrating $(\widetilde{P}(u,v)\cdot E)^2$, we obtain
$S(W_{\bullet, \bullet,\bullet}^{Y,\widetilde{S}};O)=\frac{13}{24}+F_O$.

If $O\not\in\widetilde{C}\cup\widetilde{R}$, then $F_O=0$.
Similarly, if $O\in\widetilde{C}$, then $O\not\in\widetilde{R}$, which gives

Finally, if $O\in\widetilde{R}$, then $O\not\in\widetilde{C}$, which gives

Summarizing, we get $S(W_{\bullet, \bullet,\bullet}^{\widetilde{S},E};O)\leqslant\frac{5}{6}$ for every point $O\in E$,
which contradicts (<ref>), because $S_X(S)=\frac{5}{12}$ and $S(W_{\bullet,\bullet}^{\widetilde{S}};E)=\frac{17}{12}<2$.
This shows that $X$ is K-stable.

All smooth Fano threefolds in the family №2.6 are K-stable.

§ FAMILY №2.7

Now, let us fix three smooth quadric hypersurfaces  $\mathscr{Q}$, $\mathscr{Q}^\prime$ and $\mathscr{Q}^{\prime\prime}$ in $\mathbb{P}^4$
such that their intersection is a smooth curve of degree $8$ and genus $5$.
We set $\mathscr{C}=\mathscr{Q}\cap\mathscr{Q}^\prime\cap\mathscr{Q}^{\prime\prime}$.
Let $\pi\colon X\to \mathscr{Q}$ be the blow up of the smooth curve $\mathscr{C}$.
Then $X$ is a smooth Fano threefold, which is contained in the family №2.7.
Moreover, all smooth Fano threefolds in this family can be obtained in this way.
Note that $-K_X^3=14$ and $\mathrm{Aut}(X)$ is finite [9].

The pencil generated by the surfaces $\mathscr{Q}^\prime\vert_{\mathscr{Q}}$ and $\mathscr{Q}^{\prime\prime}\vert_{\mathscr{Q}}$
gives a rational map $\mathscr{Q}\dasharrow\mathbb{P}^1$, which fits the following commutative diagram:

\xymatrix{
\mathscr{Q}\ar@{-->}[rr]&&\mathbb{P}^1}

where $\phi$ is a fibration into quartic del Pezzo surfaces.
Let $E$ be the $\pi$-exceptional surface,
and let $H=\pi^*(\mathcal{O}_{\mathbb{P}^4}(1)\vert_{\mathscr{Q}})$.
Then $-K_X\sim 3H-E$, the morphism $\phi$ is given by the linear system $|2H-E|$, and $E\cong\mathscr{C}\times\mathbb{P}^1$.

Let $S$ be a fiber of the morphism $\phi$. Then $S$ has at most Du Val singularities.

Since $E\vert_{S}$ is a smooth curve, the surface $S$ is smooth along $E\vert_{S}$,
which implies that it has at most isolated singularities. Hence, we conclude that $S$ is normal and irreducible.

Note that $S\cong\pi(S)$. Suppose that the singularities of the surface $\pi(S)$ are not Du Val.
Then it follows from <cit.> that $\pi(S)$ is a cone in $\mathbb{P}^4$ over a quartic elliptic curve.
Let $P$ be the vertex of the cone $\pi(S)$, and let $T_P$ be the hyperplane section of the quadric hypersurface $\mathscr{Q}$ that is singular at $P$.
Then $T_P$ contains all lines in $\mathscr{Q}$ that pass through $P$,
which implies that $T_P$ contains $\pi(S)$.
This is impossible, since $T_P$ is a quadric cone.

The goal of this section is to prove that $X$ is K-stable.
To do this, we fix a point $P\in X$.
By [14, 16], to prove that $X$ is K-stable, it is enough to show that $\delta_P(X)>1$.

Let $S$ be the fiber of the morphism $\phi$ containing $P$.
Then $S$ is a quartic del Pezzo surface, and $S$ has at most Du Val singularity at $P$ by Lemma <ref>.
Moreover, if $S$ is singular at $P$,
then  $P$ is a singular point of the surface $S$
of type $\mathbb{A}_1$, $\mathbb{A}_2$, $\mathbb{A}_3$, $\mathbb{A}_4$, $\mathbb{D}_4$ or $\mathbb{D}_5$, see [11].

If $\delta_P(S)>\frac{54}{55}$ or $P\in\mathrm{Sing}(S)$ and $\delta_P(S)>\frac{27}{28}$, then $\delta_P(X)>1$.

Take $u\in\mathbb{R}_{\geqslant 0}$. Since $S\sim 2H-E$, we have

-K_X-uS\sim_{\mathbb{R}} (3-2u)H-(u-1)E\sim_{\mathbb{R}}\Big(\frac{3}{2}-u\Big)S+\frac{1}{2}E.

Using this, we conclude that the divisor $-K_X-uS$ is pseudoeffective if and only if $u\leqslant \frac{3}{2}$.
For $u\leqslant \frac{3}{2}$, let $P(u)$ be the positive part of Zariski decomposition of the divisor $-K_X-uS$,
and let $N(u)$  be the negative part of Zariski decomposition of the divisor $-K_X-uS$.

&-K_X-uS\ \text{if $0\leqslant u\leqslant 1$}, \\
&(3-2u)H\ \text{if $1\leqslant u\leqslant \frac{3}{2}$},
\endaligned
\right.

&0\ \text{if $0\leqslant u\leqslant 1$}, \\
&(u-1)E\ \text{if $1\leqslant u\leqslant \frac{3}{2}$}.
\endaligned
\right.

This gives

S_{X}(S)=\frac{1}{14}\int_{0}^{\frac{3}{2}}\big(P(u)\big)^3du=\frac{1}{14}\int_{0}^{1}(14-12u)du+\frac{1}{14}\int_{1}^{\frac{3}{2}} 2(3-2u)^{3}du=\frac{33}{56}<1.

Now, using <cit.> and <cit.>, we get

\begin{equation}
\label{equation:2-7-S-delta}
\delta_P(X)\geqslant \min\left\{\frac{1}{S_X(S)},\inf_{\substack{F/S\\ P\in C_S(F)}}\frac{A_S(F)}{S(W_{\bullet,\bullet}^S;F)}\right\},
\end{equation}

where the infimum is taken by all prime divisors $F$ over the surface $S$ such that $P\in C_S(F)$,
and $S(W_{\bullet,\bullet}^S;F)$ can be computed using <cit.> as follows:

S\big(W_{\bullet,\bullet}^S; F\big)=\frac{3}{14}\int_0^{\frac{3}{2}}\big(P(u)\big\vert_{S}\big)^2\mathrm{ord}_{F}\big(N(u)\big\vert_{S}\big)du+\frac{3}{14}\int_{0}^{\frac{3}{2}}\int_0^\infty \mathrm{vol}\big(P(u)\big\vert_{S}-vF\big)dvdu.

Let $E_S=E|_S$. Then $E_S\in |-2K_S|$ and $E_S\cong\mathscr{C}$. Moreover, the surface $S$ is smooth in a neighborhood of the curve $E_S$.
Furthermore, we have

&-K_S\ \text{if $0\leqslant u\leqslant 1$}, \\
&(3-2u)(-K_S)\ \text{if $1\leqslant u\leqslant \frac{3}{2}$},
\endaligned
\right.

&0\ \text{if $0\leqslant u\leqslant 1$}, \\
&(u-1)E_S\ \text{if $1\leqslant u\leqslant \frac{3}{2}$}.
\endaligned
\right.

Let $F$ be any prime divisor over $S$ such that $P\in C_S(F)$. Then

\begin{multline*}
\quad S\big(W_{\bullet,\bullet}^S; F\big)=\frac{3}{14}\int_{1}^{\frac{3}{2}}4(u-1)(3-2u)^{2}\mathrm{ord}_{F}\big(E_{S}\big)du+\\
+\frac{3}{14}\int_{0}^{1} \int_{0}^{\infty}\mathrm{vol}\big(-K_S-vF\big)dvdu+\\
\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+\frac{3}{14}\int_{1}^{\frac{3}{2}}\int_{0}^{\infty}\mathrm{vol}\big((3-2u)(-K_S)-vF\big)dvdu=\\
\end{multline*}

because log pair $(S,E_S)$ is log canonical. Thus, if  $\delta_P(S)>\frac{54}{55}$, then (<ref>) gives $\delta_P(X)>1$.
Similarly, if $P\in\mathrm{Sing}(S)$, then $P\not\in E_S$, so that $\mathrm{ord}_{F}(E_{S})=0$, which implies that

S\big(W_{\bullet,\bullet}^S; F\big)=\frac{27}{28}S_S(F)\leqslant\frac{27A_{S}(F)}{28\delta_{P}(S)}.

Hence, in this case, it follows from (<ref>) that $\delta_P(X)>1$ provided that $\delta_P(S)>\frac{27}{28}$.

If $S$ is smooth, then $\delta_P(X)>1$.

If $S$ is smooth, then $\delta(S)=\frac{4}{3}$ by <cit.>. Now apply Lemma <ref>.

If $P$ is not contained in a line in $S$, then $\delta_P(X)>1$.

If $P$ is not contained in a line in $S$, then $P$ is a smooth point of the surface $S$,
and the blow up of the surface $S$ at this point is a (possibly singular) cubic surface in $\mathbb{P}^3$.
Thus, arguing exactly as in the end of the proof of <cit.>, we obtain $\delta_P(S)\geqslant \frac{3}{2}$,
which implies that $\delta_P(X)>1$ by Lemma <ref>.

Now, let $T$ be a surface in the linear system $|H|$ such that $P\in T$, and let $\mathcal{Q}=\pi(T)$.
Then $\mathcal{Q}$ is a hyperplane section of the hypersurface $\mathscr{Q}$, so both $\mathcal{Q}$ and $T$ are irreducible.
In the following, we will choose $T$ such that the surface $\mathcal{Q}$ is smooth, so that $\mathcal{Q}\cong\mathbb{P}^1\times\mathbb{P}^1$.

Suppose that $T$ is a general surface in the linear system $|H|$ such that $P\in T$.
Then the (scheme) intersection $S\cap T$ is an irreducible reduced curve.

Let $\rho\colon\widetilde{S}\to S$ be a blow up of the quartic del Pezzo surface $S$ at the point $P$,
and let $Z$ be the proper transform of the curve $T\vert_{S}$ on the surface $\widetilde{S}$.
Then $|Z|$ has no base points and gives the morphism $\eta\colon\widetilde{S}\to\mathbb{P}^3$
that fits the following commutative diagram:

\xymatrix{

where $S\dasharrow\mathbb{P}^3$ is a projection from $P$. Moreover, if $P$ is a smooth point of the surface $S$,
then $Z^2=3$, and the image of the morphism $\eta$ is an irreducible cubic surface in $\mathbb{P}^3$.
Similarly, if $P$ is a singular point of the surface $S$, then we have ,
and the image of the morphism $\eta$ is an irreducible quadric surface.
Therefore, we conclude that the curve $Z$ must be irreducible and reduced (by Bertini theorem),
which implies that the intersection $S\cap T$ is also irreducible and reduced.

Suppose that $S$ is singular at $P$, and $T$ is a general surface in $|H|$ that passes through the point $P$.
Then, choosing appropriate coordinates $[x:y:z:t:w]$ on $\mathbb{P}^4$,
we may assume that $\pi(P)=[0:0:0:0:1]$, and the surface $\pi(S)$ is given in $\mathbb{P}^4$ by

\left\{\aligned
&at^2+btx+f_2(x,y,z)=0, \\
&wt=g_2(x,y,z), \\
\endaligned
\right.

where $a$ and $b$ are complex numbers, $f_2(x,y,z)$ and $g_2(x,y,z)$ are non-zero quadratic homogeneous polynomials.
In the chart $w\ne 0$, the surface $\pi(S)$ is given by

\left\{\aligned
&at^2+btx+f_2(x,y,z)=0, \\
&t=g_2(x,y,z), \\
\endaligned
\right.

where now we consider $x$, $y$, $z$, $t$ as affine coordinates on $\mathbb{C}^4$.
Then $\pi(S)\cap\mathcal{Q}$ is cut out on the surface $\pi(S)$ by $c_1x+c_2y+c_3z+c_4t=0$,
where $c_1$, $c_2$, $c_3$, $c_4$ are general numbers.
The affine part of the surface $\pi(S)$ is isomorphic to the hypersurface in $\mathbb{C}^3$ given by

and the affine part of the curve $\pi(S)\cap\mathcal{Q}$ is cut out by .
If $P$ is a singular point of the surface $S$ of type $\mathbb{D}_4$ or $\mathbb{D}_5$,
then $S\cap T$ has an ordinary cusp at the point $P$, which easily implies that the intersection $S\cap T$ is reduced and irreducible.
Similarly, if $P$ is a Du Val singular point of the surface $S$ of type $\mathbb{A}_1$, $\mathbb{A}_2$, $\mathbb{A}_3$ or $\mathbb{A}_4$,
then  the intersection $S\cap T$ has an isolated ordinary double singularity at $P$.

Observe that the morphism $\pi\colon X\to\mathscr{Q}$ induces a birational morphism $\varpi\colon T\to \mathcal{Q}$,
and the morphism $\phi\colon X\to\mathbb{P}^1$ induces a fibration $\varphi\colon T\to\mathbb{P}^1$ that both fit the following commutative diagram:

\xymatrix{
\mathcal{Q}\ar@{-->}[rr]&&\mathbb{P}^1}

where $\mathcal{Q}\dasharrow\mathbb{P}^1$ is a map given by the pencil generated by the curves
$\mathscr{Q}^\prime\vert_{\mathcal{Q}}$ and $\mathscr{Q}^{\prime\prime}\vert_{\mathcal{Q}}$.
In the following, we will always choose $T\in|H|$ such that the quadric surface $\mathcal{Q}$ is smooth,
and $T$ is either smooth or has one isolated ordinary singularity,
which would imply that a general fiber of the induced fibration $\varphi\colon T\to\mathbb{P}^1$ is a smooth elliptic curve.
Let $\mathcal{C}=S\vert_{T}$.
Then $\mathcal{C}$ is the fiber of the (elliptic) fibration $\varphi$ that contains the point $P$.

Let $u$ be a non-negative real number.
Then $-K_X-uT\sim_{\mathbb{R}}(3-u)H-E\sim_{\mathbb{R}}(1-u)H+S$,
which implies that $-K_X-uT$ is nef $\iff$ $-K_X-uT$ is pseudoeffective $\iff$ $u\in[0,1]$.
Integrating, we get $S_{X}(T)=\frac{9}{28}<1$. For simplicity, we let $P(u)=-K_X-uT$.

Suppose that $S$ is singular at $P$. Then $\delta_P(X)>1$.

Now, let us choose $T\in|H|$ such that $T$ is a general surface in $|H|$ that contains $P$.
Then $T$ and $\mathcal{Q}$ are smooth, and $\varpi$ is a blow up of the eight intersection points $\mathscr{C}\cap \mathcal{Q}$.
Moreover, by Lemma <ref>, the curve $\mathcal{C}$ is an irreducible singular curve of arithmetic genus $1$.
Thus, we have $P=\mathrm{Sing}(\mathcal{C})$. Furthermore, using Remark <ref>, we see that

* either $\mathcal{C}$ has an isolated ordinary double singularity at $P$,

* or the curve $\mathcal{C}$ has an ordinary cusp at the point $P$.

Recall that $\mathcal{Q}$ is a smooth quadric surface, so that it contains exactly two lines that pass though $\pi(P)$.
Since $T$ is chosen to be general, these lines are disjoint from   $\mathscr{C}\cap \mathcal{Q}$.
Denote by $L_1$ and $L_2$ the proper transforms of these lines on $T$. Then

Now, we let $\sigma\colon\widetilde{T}\to T$ be the blow up of the point $P$,
we let $\mathbf{e}$ be the $\sigma$-exceptional curve,
and we denote by $\widetilde{\mathcal{C}}$, $\widetilde{L}_1$, $\widetilde{L}_2$ the proper transforms on $\widetilde{T}$ of the curves  $\mathcal{C}$, $L_1$, $L_2$,
respectively. Take a non-negative real number $v$. Then

\begin{equation}
\label{equation:2-7-C-singular-blow-up}
\sigma^*\big(P(u)\big\vert_{T}\big)-v\mathbf{e}\sim_{\mathbb{R}}\widetilde{\mathcal{C}}+(1-u)(\widetilde{L}_{1}+\widetilde{L}_{2})+(4-2u-v)\mathbf{e}.
\end{equation}

Note that the curves $\widetilde{\mathcal{C}}$, $\widetilde{L}_1$, $\widetilde{L}_2$ are disjoint.
Moreover, we have $\widetilde{L}_{1}^2=\widetilde{L}_{2}^2=-1$ and $\widetilde{\mathcal{C}}^2=-4$.
Thus, using (<ref>), we see that $\sigma^*(P(u)\vert_{T})-v\mathbf{e}$ is pseudoeffective $\iff$ $v\leqslant 4-2u$.

Let $\widetilde{P}(u,v)$ be the positive part of Zariski decomposition of the divisor $\sigma^*(P(u)\vert_{T})-v\mathbf{e}$,
and let $\widetilde{N}(u,v)$ be its negative part. Then it follows from <cit.> that

\begin{equation}
\label{equation:2-7-C-singular-node}
\delta_P(X)\geqslant\min\left\{\frac{1}{S_X(T)},
\frac{2}{S(W_{\bullet,\bullet}^{\widetilde{T}};\mathbf{e})},\inf_{O\in \mathbf{e}}\frac{1}{S(W_{\bullet, \bullet,\bullet}^{\widetilde{T},\mathbf{e}};O)}\right\},
\end{equation}

where $S(W_{\bullet,\bullet}^{\widetilde{T}};\mathbf{e})$ and $S(W_{\bullet, \bullet,\bullet}^{\widetilde{T},\mathbf{e}};O)$ are defined  in <cit.>,
and these two numbers can be computed using formulas described in <cit.>.
Namely, we have

S\big(W_{\bullet, \bullet,\bullet}^{\widetilde{T},\mathbf{e}};O\big)=\frac{3}{14}\int_0^1\int_0^{4-2u}\Big(\big(\widetilde{P}(u,v)\cdot\mathbf{e}\big)\Big)^2dvdu+\frac{3}{7}\int_0^1\int_0^{4-2u}\big(\widetilde{P}(u,v)\cdot\mathbf{e}\big)\mathrm{ord}_O\Big(\widetilde{N}(u,v)\big|_{\mathbf{e}}\Big)dvdu,

where $O$ is a point in $\mathbf{e}$.
Moreover, using (<ref>), we compute $\widetilde{P}(u,v)$ and $\widetilde{N}(u,v)$ as follows:

\widetilde{P}(u,v)=\left\{\aligned
&\widetilde{\mathcal{C}}+(1-u)(\widetilde{L}_{1}+\widetilde{L}_{2})+(4-2u-v)\mathbf{e}\ \text{if $0\leqslant v\leqslant 2-2u$}, \\
&\frac{4-2u-v}{2}\widetilde{\mathcal{C}}+(1-u)(\widetilde{L}_{1}+\widetilde{L}_{2})+(4-2u-v)\mathbf{e}\ \text{if $2-2u\leqslant v\leqslant 3-u$}, \\
&\frac{4-2u-v}{2}\big(\widetilde{\mathcal{C}}+2(\widetilde{L}_{1}+\widetilde{L}_{2})+2\mathbf{e}\big)\ \text{if $3-u\leqslant v\leqslant 4-2u$}, \\
\endaligned
\right.

\widetilde{N}(u,v)=\left\{\aligned
&0\ \text{if $0\leqslant v\leqslant 2-2u$}, \\
&\frac{v-2+2u}{2}\widetilde{\mathcal{C}}\ \text{if $2-2u\leqslant v\leqslant 3-u$}, \\
&\frac{v-2+2u}{2}\widetilde{\mathcal{C}}+(v-3+u)(\widetilde{L}_{1}+\widetilde{L}_{2}) \ \text{if $3-u\leqslant v\leqslant 4-2u$}.\\
\endaligned
\right.

Thus, we have

\big(\widetilde{P}(u,v)\big)^2=\left\{\aligned
&2(1-u)(5-u)-v^{2}\ \text{if $0\leqslant v\leqslant 2-2u$}, \\
&2(1-u)(7-3u-2v)\ \text{if $2-2u\leqslant v\leqslant 3-u$}, \\
&2(4-2 u-v)^{2}\ \text{if $3-u\leqslant v\leqslant 4-2u$}, \\
\endaligned
\right.

\widetilde{P}(u,v)\cdot\mathbf{e}=\left\{\aligned
&v\ \text{if $0\leqslant v\leqslant 2-2u$}, \\
&2(1-u)\ \text{if $2-2u\leqslant v\leqslant 3-u$}, \\
&2(4-2u-v)\ \text{if $3-u\leqslant v\leqslant 4-2u$}. \\
\endaligned
\right.

Now, integrating, we get $S(W_{\bullet,\bullet}^{\widetilde{T}};\mathbf{e})=\frac{51}{28}<2$.

Let $O$ be an arbitrary point in $\mathbf{e}$.
If $O\not\in\widetilde{L}_1\cup\widetilde{L}_2\cup\widetilde{\mathcal{C}}$, then we compute $S(W_{\bullet, \bullet,\bullet}^{\widetilde{T},\mathbf{e}};O)=\frac{4}{7}$.
Similarly, if $O\in\widetilde{L}_1\cup\widetilde{L}_2$, then $S(W_{\bullet, \bullet,\bullet}^{\widetilde{T},\mathbf{e}};O)=\frac{17}{28}$.
Finally, if $O\in\widetilde{\mathcal{C}}$, then

\begin{multline*}
S\big(W_{\bullet, \bullet,\bullet}^{\widetilde{T},\mathbf{e}};O\big)=\frac{4}{7}+\frac{3}{7}\int_{0}^{1}\int_{2-2u}^{3-u} 2(1-u)\frac{v-2+2 u}{2}\mathrm{ord}_O\big(\widetilde{\mathcal{C}}\big\vert_{\mathbf{e}}\big)dvdu+\\
+\frac{3}{7}\int_{0}^{1}\int_{3-u}^{4-2u} 2(4-2u-v)\frac{v-2+2 u}{2}\mathrm{ord}_O\big(\widetilde{\mathcal{C}}\big\vert_{\mathbf{e}}\big)dvdu=\frac{4}{7}+\frac{17}{56}\mathrm{ord}_O\big(\widetilde{\mathcal{C}}\big\vert_{\mathbf{e}}\big).
\end{multline*}

Hence, if $\widetilde{\mathcal{C}}$ intersects $\mathbf{e}$ transversally, then $S(W_{\bullet, \bullet,\bullet}^{\widetilde{T},\mathbf{e}};O)<1$, so that $\delta_P(X)>1$ by (<ref>).

Therefore, to complete the proof of the lemma, we may assume that $\widetilde{\mathcal{C}}$ is tangent to $\mathbf{e}$.
This means that $\mathcal{C}$ has a cusp at $P$, and the intersection $\widetilde{\mathcal{C}}\cap\mathbf{e}$ consists of a single point.

Now, as in the proof of Lemma <ref>, we consider the following commutative diagram:

\xymatrix{
\widetilde{T}\ar@{->}[d]_{\sigma}&&\overline{T}\ar@{->}[ll]_{\gamma}\ar@{->}[d]^{\upsilon}\\%
T && \widehat{T}\ar@{->}[ll]_{\varsigma}}

where $\gamma$ is a composition of the blow up of the point $\widetilde{\mathcal{C}}\cap\mathbf{e}$
with the blow up of the unique intersection point of the proper transforms of the curves $\widetilde{\mathcal{C}}$ and $\mathbf{e}$,
$\upsilon$ is the birational contraction of all $(\sigma\circ\gamma)$-exceptional curves that are not $(-1)$-curve,
and $\varsigma$ is the birational contraction of the proper transform of the unique $\gamma$-exceptional curve that is $(-1)$-curve.
Then $\widehat{T}$ has two singular points:

* a cyclic quotient singularity of type $\frac{1}{2}(1,1)$, which we denote by $O_2$;

* a cyclic quotient singularity of type $\frac{1}{3}(1,1)$, which we denote by $O_3$.

Let $\mathbf{f}$ be the $\varsigma$-exceptional curve,
let $\widehat{\mathcal{C}}$ be the proper transform on $\widehat{T}$ of the curve $\mathcal{C}$,
and let $\widehat{L}_1$ and $\widehat{L}_2$ be the proper transforms on $\widehat{T}$ of the curves $L_1$ and $L_2$, respectively.
Then the curves $\mathbf{f}$, $\widehat{\mathcal{C}}$, $\widehat{L}_1$, $\widehat{L}_2$ are smooth,
and the curve $\mathbf{f}$ contains both points $O_2$ and $O_3$, which are not contained in  $\widehat{\mathcal{C}}$.
Moreover, we have

\widehat{L}_1\cap\widehat{L}_2=\widehat{L}_1\cap\mathbf{f}=\widehat{L}_2\cap\mathbf{f}=O_3.

Furthermore, we have $A_T(\mathbf{f})=5$, $\varsigma^*(\mathcal{C})\sim\widehat{\mathcal{C}}+6\mathbf{f}$,
$\varsigma^*(L_1)\sim\widehat{L}_1+2\mathbf{f}$, $\varsigma^*(L_2)\sim\widehat{L}_2+2\mathbf{f}$,
and the intersection form of the curves $\mathbf{f}$, $\widehat{\mathcal{C}}$, $\widehat{L}_1$ and $\widehat{L}_2$ is given in the following table:

$\mathbf{f}$     $\widehat{\mathcal{C}}$     $\widehat{L}_1$     $\widehat{L}_2$

$\mathbf{f}$                 $-\frac{1}{6}$     $1$      $\frac{1}{3}$     $\frac{1}{3}$

$\widehat{\mathcal{C}}$      $1$                 $-6$     $0$     $0$

$\widehat{L}_1$              $\frac{1}{3}$       $0$     $-\frac{2}{3}$     $\frac{1}{3}$

$\widehat{L}_2$              $\frac{1}{3}$       $0$     $\frac{1}{3}$     $-\frac{2}{3}$

For a non-negative real number $v$, we have

\varsigma^*\big(P(u)\big\vert_{T}\big)-v\mathbf{f}\sim_{\mathbb{R}}\widetilde{\mathcal{C}}+(1-u)(\widetilde{L}_{1}+\widetilde{L}_{2})+(10-4u-v)\mathbf{f},

which implies that the divisor $\varsigma^*(P(u)\vert_{T})-v\mathbf{f}$ is pseudoeffective if and only if $v\leqslant 10-4u$,
because the intersection form of the curves  $\widehat{\mathcal{C}}$, $\widehat{L}_1$, $\widehat{L}_2$ is negative definite.

Let $\widehat{P}(u,v)$ be the positive part of Zariski decomposition of the divisor $\varsigma^*(P(u)\vert_{T})-v\mathbf{f}$,
and let $\widehat{N}(u,v)$ be its negative part.
Set $\Delta_\mathbf{f}=\frac{1}{2}O_2+\frac{2}{3}O_3$.
By <cit.>, we get

\begin{equation}
\label{equation:2-7-C-singular-cusp}
\delta_P(X)\geqslant\min\left\{\frac{1}{S_X(T)},
\frac{5}{S(W_{\bullet,\bullet}^{\widehat{T}};\mathbf{f})},\inf_{O\in \mathbf{f}}\frac{1-\mathrm{ord}_O\big(\Delta_\mathbf{f}\big)}{S(W_{\bullet, \bullet,\bullet}^{\widehat{T},\mathbf{f}};O)}\right\},
\end{equation}

where $S(W_{\bullet,\bullet}^{\widehat{T}};\mathbf{f})$ and $S(W_{\bullet, \bullet,\bullet}^{\widehat{T},\mathbf{f}};O)$ are defined as $S(W_{\bullet,\bullet}^{\widetilde{T}};\mathbf{e})$ and $S(W_{\bullet, \bullet,\bullet}^{\widetilde{T},\mathbf{e}};O)$ used earlier.
Moreover, it follows from <cit.> that

S\big(W_{\bullet, \bullet,\bullet}^{\widehat{T},\mathbf{f}};O\big)=\frac{3}{14}\int_0^1\int_0^{10-4u}\Big(\big(\widehat{P}(u,v)\cdot\mathbf{f}\big)\Big)^2dvdu+\frac{3}{7}\int_0^1\int_0^{10-4u}\big(\widehat{P}(u,v)\cdot\mathbf{f}\big)\mathrm{ord}_O\Big(\widehat{N}(u,v)\big|_{\mathbf{f}}\Big)dvdu.

Moreover, we compute $\widehat{P}(u,v)$ and $\widehat{N}(u,v)$ as follows:

\widehat{P}(u,v)=\left\{\aligned
&\widetilde{\mathcal{C}}+(1-u)(\widetilde{L}_{1}+\widetilde{L}_{2})+(10-4u-v)\mathbf{f}\ \text{if $0\leqslant v\leqslant 4-4u$}, \\
&\frac{10-4u-v}{6}\widetilde{\mathcal{C}}+(1-u)(\widetilde{L}_{1}+\widetilde{L}_{2})+(10-4u-v)\mathbf{f}\ \text{if $4-4u\leqslant v\leqslant 9-3u$}, \\
&\frac{10-4u-v}{6}\big(\widehat{\mathcal{C}}+6(\widehat{L}_{1}+\widehat{L}_{2})+6\mathbf{f}\big)\ \text{if $9-3u\leqslant v\leqslant 10-4u$}, \\
\endaligned
\right.

\widehat{N}(u,v)=\left\{\aligned
&0\ \text{if $0\leqslant v\leqslant 4-4u$}, \\
&\frac{v-4+4u}{6}\widehat{\mathcal{C}}\ \text{if $4-4u\leqslant v\leqslant 9-3u$}, \\
&\frac{v-4+4u}{6}\widehat{\mathcal{C}}+(v-9+3u)(\widetilde{L}_{1}+\widetilde{L}_{2})\ \text{if $9-3u\leqslant v\leqslant 10-4u$}. \\
\endaligned
\right.

This gives

\big(\widehat{P}(u,v)\big)^2=\left\{\aligned
&2(1-u)(5-u)-\frac{v^{2}}{6}\ \text{if $0\leqslant v\leqslant 4-4u$}, \\
&\frac{2(1-u)(19-7u-2v)}{3}\ \text{if $4-4u\leqslant v\leqslant 9-3u$}, \\
&\frac{2(10-4u-v)^{2}}{3}\ \text{if $9-3u\leqslant v\leqslant 10-4u$}, \\
\endaligned
\right.

\widehat{P}(u,v)\cdot\mathbf{f}=\left\{\aligned
&\frac{v}{6}\ \text{if $0\leqslant v\leqslant 4-4u$}, \\
&\frac{2(1-u)}{3}\ \text{if $4-4u\leqslant v\leqslant 9-3u$}, \\
&\frac{2(10-4 u-v)}{3}\ \text{if $9-3u\leqslant v\leqslant 10-4u$}. \\
\endaligned
\right.

Now, integrating, we get $S(W_{\bullet,\bullet}^{\widehat{T}};\mathbf{f})=\frac{135}{28}<A_{T}(\mathbf{f})=5$.

Let $O$ be a point in $\mathbf{f}$.
If $O\not\in\widehat{L}_1\cup\widehat{L}_2\cup\widehat{\mathcal{C}}$, then

\begin{multline*}
\quad\quad\quad S\big(W_{\bullet, \bullet,\bullet}^{\widehat{T},\mathbf{f}};O\big)=\frac{3}{14}\int_{0}^{1}\int_{0}^{4-4u}\left(\frac{v}{6}\right)^{2}dvdu+\frac{3}{14}\int_{0}^{1}\int_{4-4u}^{9-3u}\left(\frac{2(1-u)}{3}\right)^{2}dvdu+\\
+\frac{3}{14}\int_{0}^{1}\int_{9-3u}^{10-4 u}\left(\frac{2(10-4 u-v)}{3}\right)^{2}dvdu=\frac{13}{63}.\quad\quad\quad
\end{multline*}

Similarly, if $O=\mathbf{f}\cap\widetilde{\mathcal{C}}$, then

\begin{multline*}
\quad\quad\quad S\big(W_{\bullet, \bullet,\bullet}^{\widehat{T},\mathbf{f}};O\big)=\frac{13}{63}+
\frac{3}{7}\int_{0}^{1}\int_{4-4u}^{9-3u}\frac{2(1-u)}{3}\times\frac{v-4+4 u}{6}dvdu+\\
+\frac{3}{7}\int_{0}^{1}\int_{9-3u}^{10-4u}\frac{2(10-4 u-v)}{3}\times\frac{v-4+4u}{6}dvdu=\frac{13}{63}+\frac{193}{504}=\frac{33}{56}.
\end{multline*}

Likewise, if $O=O_3$, we compute $S(W_{\bullet, \bullet,\bullet}^{\widehat{T},\mathbf{f}};O)=\frac{3}{14}$.
So, using (<ref>), we get $\delta_P(X)>1$.

Thus, to prove that $\delta_P(X)>1$, we may assume that $S$ is singular, but $P\not\in\mathrm{Sing}(S)$.

Suppose that $P\not\in E$. Then $\delta_P(X)>1$.

By Corollary <ref>, we may assume that $S$ contains a line $L$ that passes through $P$.
Then $\pi(L)$ is a line in $\mathscr{Q}$. Note that $\pi(L)\cap\mathscr{C}\ne\varnothing$, and one of the following cases holds:

Case 1: the line $\pi(L)$ intersects the curve $\mathscr{C}$ transversally at $2$ points,

Case 2: the line $\pi(L)$ is tangent to the curve $\mathscr{C}$ at their single intersection point.

Now, let us choose $T$ to be a sufficiently general surface in $|H|$ that passes through $L$.
If the intersection $\pi(L)\cap\mathscr{C}$ consists of two points, then $\varpi\colon T\to \mathcal{Q}$ is a blow up of eight distinct points of the transversal intersection $\mathcal{Q}\cap\mathscr{C}$,
which implies that  $T$ is smooth.
On the other hand, if $L\cap\mathscr{C}$ consists of one point, then $T$ has one ordinary double point, which is not contained in the curve $L$.
We have $\mathcal{C}=S\vert_{T}=L+Z$,
where $Z$ is a smooth rational irreducible curve such that $\pi(Z)$ is a smooth twisted cubic in $\mathscr{Q}$ that intersects the curve $\mathscr{C}$
transversally by six distinct points, which we denote by $Q_3,Q_4,Q_5,Q_6,Q_7,Q_8$.
Moreover, if  $\pi(L)\cap\mathscr{C}$ consists of two distinct points, we denote these points by $Q_1$ and $Q_2$.
Likewise, if $\pi(L)\cap\mathscr{C}$ consists of one point, we let $Q_1=Q_2=\pi(Z)\cap \mathscr{C}$. Then

Case 1: the morphism $\varpi\colon T\to\mathcal{Q}$ is the blow up of the points $Q_1,Q_2,\ldots,Q_8$,

Case 2:  the morphism $\varpi\colon T\to\mathcal{Q}$  is a composition of the blow up of the points $Q_3,\ldots,Q_8$
with a weighted blow up with weights $(1,2)$ of the point $Q_1=Q_2$.

Since $T$ is a general surface in $|H|$ that contains the line $L$, we may assume that $P\not\in Z$.
Likewise, we may assume further that $Z$ is contained in the smooth locus of the surface $T$.
Moreover, we may also assume that the quadric surface $\mathcal{Q}$ does not contain lines that pass through one point in the set $\{Q_1,Q_2,\pi(P)\}$
and one point in $\{Q_3,Q_4,Q_5,Q_6,Q_7,Q_8\}$.
Indeed, let $\mathcal{Q}^\prime$, $\mathcal{Q}^{\prime\prime}$, $\mathcal{Q}^{\prime\prime\prime}$ be the hyperplane sections of the quadric $\mathscr{Q}$
that are singular at the points  $Q_1$, $Q_2$, $\pi(P)$, respectively.
Then $\mathcal{Q}^\prime$, $\mathcal{Q}^{\prime\prime}$, $\mathcal{Q}^{\prime\prime\prime}$ are cones,
and every line in $\mathscr{Q}$ containing a point in $\{Q_1,Q_2,\pi(P)\}$ is a ruling of one of these cones.
On the other hand, we have $\mathscr{C}\not\subset\mathcal{Q}^\prime\cup\mathcal{Q}^{\prime\prime}\cup\mathcal{Q}^{\prime\prime\prime}$,
because $\mathscr{C}$ is not contained in a hyperplane.
This implies that the quadric threefold $\mathscr{Q}$ contains at most finitely many lines that pass through a point in $\{Q_1,Q_2,\pi(P)\}$ and a point in $\mathscr{C}\setminus\{Q_1,Q_2\}$.
Therefore, we can choose the surface $T\in|H|$ such that $L\subset T$, but $\mathcal{Q}=\pi(T)$ does not contain any of these lines.

Let us identify $\mathcal{Q}=\mathbb{P}^1\times\mathbb{P}^1$ such that the line $\pi(L)$ is a curve in $\mathcal{Q}$ of degree $(0,1)$.
Denote by $\mathbf{e}_1,\ldots,\mathbf{e}_8$ the $\varpi$-exceptional curves such that $\varpi(\mathbf{e}_1)=Q_1,\ldots,\varpi(\mathbf{e}_8)=Q_8$.
Let $\mathbf{g}_3,\ldots,\mathbf{g}_8$ be the strict transforms on $T$ of the curves in $\mathcal{Q}$ of degree $(1,0)$
that pass through the points $Q_3,\ldots,Q_8$, respectively.
Then $L,Z,\mathbf{e}_1,\ldots,\mathbf{e}_8,\mathbf{g}_3,\ldots,\mathbf{g}_8$ are smooth irreducible rational curves.
In Case 1, their intersections are given in the following table:

$L$       $Z$       $\mathbf{e}_1$      $\mathbf{e}_2$       $\mathbf{e}_3$       $\mathbf{e}_4$       $\mathbf{e}_5$      $\mathbf{e}_6$       $\mathbf{e}_7$       $\mathbf{e}_8$
$\mathbf{g}_3$      $\mathbf{g}_4$      $\mathbf{g}_5$      $\mathbf{g}_6$      $\mathbf{g}_7$      $\mathbf{g}_8$

$L$                  $-2$     $2$     $1$     $1$     $0$     $0$     $0$     $0$     $0$     $0$     $1$     $1$     $1$     $1$     $1$     $1$

$Z$                  $2$     $-2$     $0$     $0$     $1$    $1$    $1$    $1$    $1$    $1$     $0$    $0$    $0$    $0$    $0$    $0$

$\mathbf{e}_1$      $1$     $0$    $-1$     $0$    $0$    $0$    $0$    $0$    $0$    $0$    $0$    $0$    $0$    $0$    $0$    $0$

$\mathbf{e}_2$      $1$     $0$    $0$     $-1$    $0$    $0$    $0$    $0$    $0$    $0$    $0$    $0$    $0$    $0$    $0$    $0$

$\mathbf{e}_3$       $0$     $1$    $0$     $0$    $-1$     $0$    $0$    $0$    $0$    $0$    $1$    $0$    $0$    $0$    $0$    $0$

$\mathbf{e}_4$       $0$    $1$     $0$    $0$     $0$    $-1$    $0$    $0$    $0$    $0$    $0$    $1$    $0$    $0$    $0$    $0$

$\mathbf{e}_5$       $0$    $1$    $0$    $0$     $0$    $0$    $-1$    $0$    $0$    $0$    $0$    $0$    $1$    $0$    $0$    $0$

$\mathbf{e}_6$       $0$    $1$   $0$     $0$    $0$    $0$    $0$    $-1$    $0$    $0$    $0$    $0$    $0$    $1$    $0$    $0$

$\mathbf{e}_7$       $0$    $1$    $0$    $0$    $0$    $0$    $0$    $0$    $-1$    $0$    $0$    $0$    $0$    $0$    $1$    $0$

$\mathbf{e}_8$       $0$    $1$    $0$   $0$     $0$    $0$    $0$    $0$    $0$    $-1$    $0$    $0$    $0$    $0$    $0$    $1$

$\mathbf{g}_3$      $1$    $0$    $0$   $0$     $1$   $0$     $0$    $0$    $0$    $0$    $-1$    $0$   $0$     $0$    $0$    $0$

$\mathbf{g}_4$      $1$    $0$   $0$    $0$     $0$    $1$    $0$    $0$   $0$     $0$    $0$    $-1$    $0$    $0$    $0$    $0$

$\mathbf{g}_5$      $1$    $0$    $0$    $0$    $0$    $0$    $1$   $0$     $0$    $0$    $0$    $0$    $-1$    $0$    $0$   $0$

$\mathbf{g}_6$      $1$    $0$    $0$   $0$     $0$    $0$    $0$    $1$    $0$   $0$     $0$    $0$    $0$    $-1$    $0$    $0$

$\mathbf{g}_7$      $1$    $0$   $0$     $0$    $0$    $0$    $0$    $0$    $1$    $0$    $0$    $0$    $0$    $0$    $-1$    $0$

$\mathbf{g}_8$      $1$    $0$    $0$    $0$    $0$    $0$    $0$   $0$     $0$   $1$     $0$    $0$    $0$    $0$    $0$    $-1$

In Case 2, we have $\mathbf{e}_1=\mathbf{e}_2$, and  $\mathbf{e}_1$ contains the singular point of $T$, so that $\mathbf{e}_1^2=-\frac{1}{2}$.
The remaining intersection numbers are exactly the same as in Case 1.

Observe that $P\not\in Z\cup\mathbf{g}_3\cup\mathbf{g}_4\cup\mathbf{g}_5\cup\mathbf{g}_6\cup\mathbf{g}_7\cup\mathbf{g}_8\cup\mathbf{e}_1\cup\mathbf{e}_2$, since $P\not\in E$ by assumption.

Recall that $P(u)=-K_X-uT$  is nef $\iff$ $P(u)$ is pseudoeffective $\iff$ $u\in[0,1]$.
Let $v$ be a non-negative real number. Then, in both Cases 1 and 2, we have

\begin{equation}
\label{equation:2-7-S-smooth-at-P-not-in-E-equivalence}
\end{equation}

which implies that the divisor $P(u)\vert_{T}-vL$ is pseudoeffective $\iff$ $v\leqslant \frac{9-5u}{4}$.

Let $P(u,v)$ be the positive part of Zariski decomposition of the divisor $P(u)\vert_{T}-vL$,
and let $N(u,v)$ be its negative part. Then it follows from <cit.> that

\begin{equation}
\label{equation:S-smooth-at-P-not-in-E-inequality}
\delta_P(X)\geqslant\min\left\{\frac{1}{S_X(T)},\frac{1}{S(W_{\bullet,\bullet}^{T};L)},\frac{1}{S(W_{\bullet, \bullet,\bullet}^{T,L};P)}\right\},
\end{equation}

S\big(W_{\bullet, \bullet,\bullet}^{T,L};P\big)=\frac{3}{14}\int_0^1\int_0^{\frac{9-5u}{4}}\Big(\big(P(u,v)\cdot L\big)\Big)^2dvdu+\frac{3}{7}\int_0^1\int_0^{\frac{9-5u}{4}}\big(P(u,v)\cdot L\big)\mathrm{ord}_P\Big(N(u,v)\big|_{L}\Big)dvdu.

Let us compute $S(W_{\bullet,\bullet}^{T};L)$ and $S(W_{\bullet, \bullet,\bullet}^{T,L};P)$.
If $0\leqslant u\leqslant \frac{1}{3}$, then, using (<ref>), we get

&\frac{9-5u-4v}{4}L+\frac{3+u}{4}Z+\frac{5-5u}{4}\big(\mathbf{e}_1+\mathbf{e}_2\big)+\frac{1-u}{4}\sum_{i=3}^8\mathbf{g}_i\ \text{if $0\leqslant v\leqslant 1$}, \\
&\frac{9-5u-4v}{4}\big(L+\mathbf{e}_1+\mathbf{e}_2\big)+\frac{3+u}{4}Z+\frac{1-u}{4}\sum_{i=3}^8\mathbf{g}_i\ \text{if $1\leqslant v\leqslant \frac{3-3u}{2}$}, \\
&\frac{9-5u-4v}{4}\big(L+Z+\mathbf{e}_1+\mathbf{e}_2\big)+\frac{1-u}{4}\sum_{i=3}^8\mathbf{g}_i\ \text{if $\frac{3-3u}{2}\leqslant v\leqslant 2-u$}, \\
&\frac{9-5 u-4 v}{4}\Big(L+Z+\mathbf{e}_1+\mathbf{e}_2+\sum_{i=3}^8\mathbf{g}_i\Big)\ \text{if $2-u\leqslant v\leqslant \frac{9-5u}{4}$}, \\
\endaligned
\right.

&0\ \text{if $0\leqslant v\leqslant 1$}, \\
&(v-1)\big(\mathbf{e}_1+\mathbf{e}_2\big) \ \text{if $1\leqslant v\leqslant \frac{3-3u}{2}$}, \\
&(v-1)\big(\mathbf{e}_1+\mathbf{e}_2\big)+\frac{2v+3u-3}{2}Z\ \text{if $\frac{3-3u}{2}\leqslant v\leqslant 2-u$}, \\
&(v-1)\big(\mathbf{e}_1+\mathbf{e}_2\big)+\frac{2v+3u-3}{2}Z+(v-2+u)\sum_{i=3}^{8}\mathbf{g}_{i}\ \text{if $2-u\leqslant v\leqslant \frac{9-5u}{4}$}, \\
\endaligned
\right.

\big(P(u,v)\big)^2=\left\{\aligned
&2u^2+(2v-12)u-2v^2-2v+10 \ \text{if $0\leqslant v\leqslant 1$}, \\
&2u^2+(2v-12)u-6v+12\ \text{if $1\leqslant v\leqslant \frac{3-3u}{2}$}, \\
&\frac{13u^2+16uv+4v^2-42u-24v+33}{2}\ \text{if $\frac{3-3u}{2}\leqslant v\leqslant 2-u$}, \\
&\frac{(9-5u-4v)^{2}}{2}\ \text{if $2-u\leqslant v\leqslant \frac{9-5u}{4}$}, \\
\endaligned
\right.

P(u,v)\cdot L=\left\{\aligned
&1-u+2v\ \text{if $0\leqslant v\leqslant 1$}, \\
&3-u\ \text{if $1\leqslant v\leqslant \frac{3-3u}{2}$}, \\
&6-4u-2v\ \text{if $\frac{3-3u}{2}\leqslant v\leqslant 2-u$}, \\
&2(9-5u-4v)\ \text{if $2-u\leqslant v\leqslant \frac{9-5u}{4}$}, \\
\endaligned
\right.

Similarly, if $\frac{1}{3}\leqslant u\leqslant 1$, then, using (<ref>), we get

&\frac{9-5u-4v}{4}L+\frac{3+u}{4}Z+\frac{5-5u}{4}\big(\mathbf{e}_1+\mathbf{e}_2\big)+\frac{1-u}{4}\sum_{i=3}^8\mathbf{g}_i\ \text{if $0\leqslant v\leqslant \frac{3-3u}{2}$}, \\
&\frac{9-5u-4v}{4}(L+Z)+\frac{5-5u}{4}\big(\mathbf{e}_1+\mathbf{e}_2\big)+\frac{1-u}{4}\sum_{i=3}^8\mathbf{g}_i\ \text{if $\frac{3-3u}{2}\leqslant v\leqslant 1$}, \\
&\frac{9-5u-4v}{4}\big(L+Z+\mathbf{e}_1+\mathbf{e}_2\big)+\frac{1-u}{4}\sum_{i=3}^8\mathbf{g}_i\ \text{if $1\leqslant v\leqslant 2-u$}, \\
&\frac{9-5 u-4 v}{4}\Big(L+Z+\mathbf{e}_1+\mathbf{e}_2+\sum_{i=3}^8\mathbf{g}_i\Big)\ \text{if $2-u\leqslant v\leqslant \frac{9-5u}{4}$}, \\
\endaligned
\right.

&0\ \text{if $0\leqslant v\leqslant \frac{3-3u}{2}$}, \\
&\frac{2v+3u-3}{2}Z\ \text{if $\frac{3-3u}{2}\leqslant v\leqslant 1$}, \\
&(v-1)\big(\mathbf{e}_1+\mathbf{e}_2\big)+\frac{2v+3u-3}{2}Z\ \text{if $1\leqslant v\leqslant 2-u$}, \\
&(v-1)\big(\mathbf{e}_1+\mathbf{e}_2\big)+\frac{2v+3u-3}{2}Z+(v-2+u)\sum_{i=3}^{8}\mathbf{g}_{i}\ \text{if $2-u\leqslant v\leqslant \frac{9-5u}{4}$}, \\
\endaligned
\right.

\big(P(u,v)\big)^2=\left\{\aligned
&2u^2+(2v-12)u-2v^2-2v+10\ \text{if $0\leqslant v\leqslant \frac{3-3u}{2}$}, \\
&\frac{(1-u)(29-13u-16v)}{2}\ \text{if $\frac{3-3u}{2}\leqslant v\leqslant 1$}, \\
&\frac{13u^2+16uv+4v^2-42u-24v+33}{2}\ \text{if $1\leqslant v\leqslant 2-u$}, \\
&\frac{(9-5u-4v)^{2}}{2}\ \text{if $2-u\leqslant v\leqslant \frac{9-5u}{4}$}, \\
\endaligned
\right.

P(u,v)\cdot L=\left\{\aligned
&1-u+2v\ \text{if $0\leqslant v\leqslant \frac{3-3u}{2}$}, \\
&4-4u\ \text{if $\frac{3-3u}{2}\leqslant v\leqslant 1$}, \\
&6-4u-2v\ \text{if $1\leqslant v\leqslant 2-u$}, \\
&2(9-5u-4v)\ \text{if $2-u\leqslant v\leqslant \frac{9-5u}{4}$}, \\
\endaligned
\right.

Therefore, we have $P\not\in\mathrm{Supp}(N(u,v))$, because $P\not\in Z\cup\mathbf{g}_3\cup\mathbf{g}_4\cup\mathbf{g}_5\cup\mathbf{g}_6\cup\mathbf{g}_7\cup\mathbf{g}_8\cup\mathbf{e}_1\cup\mathbf{e}_2$.
So, integrating $(P(u,v))^2$ and $(P(u,v)\cdot L)^2$, we get $S(W_{\bullet,\bullet}^{T};L)=\frac{423}{448}$
and $S(W_{\bullet, \bullet,\bullet}^{T,L};P)=\frac{79}{84}$, which implies that $\delta_P(X)>1$ by (<ref>).

By Lemma <ref>, to show that $\delta_P(X)>1$, we may assume that $P\in E$. Then $\pi(P)\in\mathscr{C}$.
Now, let us choose $T$ to be a sufficiently general surface in $|H|$ that contains the point $P$,
so that $\mathcal{Q}$ is a sufficiently general hyperplane section of the quadric $\mathscr{Q}$ that contains $\pi(P)$.
Then $T$ is smooth, and $\varpi\colon T\to \mathcal{Q}$ is a blow up of eight  points of the intersection $\mathcal{Q}\cap\mathscr{C}$.

Let $Q_1=\pi(P)$, let $Q_2,\ldots,Q_8$ be the remaining seven points of the intersection $\mathcal{Q}\cap\mathscr{C}$,
and let $\mathbf{e}_1,\ldots,\mathbf{e}_8$ be the $\varpi$-exceptional curves such that $\varpi(\mathbf{e}_1)=Q_1,\ldots,\varpi(\mathbf{e}_8)=Q_8$.
For every $u\in [0,1]$, we set

t(u)=\inf\Big\{v\in \mathbb R_{\geqslant 0} \ \big|\ \text{the divisor $P(u)\big|_T-v\mathbf{e}_1$ is pseudo-effective}\Big\}.

Then $t(u)$ is not very easy to compute explicitly in terms of $u\in[0,1]$. Fix $v\in [0, t(u)]$.
Let $P(u,v)$ be the positive part of the Zariski decomposition of the divisor ,
and let $N(u,v)$ be its negative part.
Then it follows from <cit.> that

\begin{equation}
\label{equation:2-7-P-in-E-inequality}
\delta_P(X)\geqslant\min\left\{\frac{1}{S_X(T)},\frac{1}{S(W_{\bullet,\bullet}^{T};\mathbf{e}_1)},\frac{1}{S(W_{\bullet, \bullet,\bullet}^{T,\mathbf{e}_1};P)}\right\},
\end{equation}

\begin{multline*}
\quad \quad \quad \quad S\big(W_{\bullet, \bullet,\bullet}^{T,\mathbf{e}_1};P\big)=\frac{3}{14}\int_0^1\int_0^{t(u)}\Big(\big(P(u,v)\cdot\mathbf{e}_1\big)\Big)^2dvdu+\\
+\frac{3}{7}\int_0^1\int_0^{t(u)}\big(P(u,v)\cdot \mathbf{e}_1\big)\mathrm{ord}_P\Big(N(u,v)\big|_{\mathbf{e}_1}\Big)dvdu.\quad \quad \quad
\end{multline*}

Let us compute $S(W_{\bullet,\bullet}^{T};\mathbf{e}_1)$ and $S(W_{\bullet, \bullet,\bullet}^{T,\mathbf{e}_1};P)$. This will take a while.

Recall that $\varphi\colon T\to\mathbb{P}^1$ is an elliptic fibration, which is given by the linear system $|-K_T|$.
As in the proof of Lemma <ref>, let us identify $\mathcal{Q}=\mathbb{P}^1\times\mathbb{P}^1$.

Let $F$ be a curve in $|-K_T|$. Then $F$ is irreducible and reduced.

Suppose that $F$ is reducible or non-reduced. Then the curve $\pi(F)$ is also reducible or non-reduced,
and every irreducible component of the curve $F$ is a smooth $(-2)$-curve.
But $\pi(F)$ is a curve in $\mathcal{Q}$ of degree $(2,2)$ that passes through the points $Q_1,Q_2,\ldots,Q_8$.
Therefore, we have one of the following possibilities:

* $\mathcal{Q}$ contains a line that passes through $Q_1$ and one point among  $Q_2,\ldots,Q_8$,

* $\mathcal{Q}$ contains a line that passes through two points among  $Q_2,\ldots,Q_8$,

* $\mathcal{Q}$ contains a conic that passes through $Q_1$ and three points among  $Q_2,\ldots,Q_8$.

Recall that $\mathcal{Q}$ is a general hyperplane section of the quadric $\mathscr{Q}$ that contains $Q_1=\pi(P)$.
As we already mentioned in the proof of Lemma <ref>,
the quadric $\mathscr{Q}$ contains finitely many lines that pass through $Q_1$ and a point in $\mathscr{C}\setminus Q_1$.
Thus, since $\mathcal{Q}$ is assumed to be general, the quadric $\mathcal{Q}$ does not contain any of these lines,
so that $\mathcal{Q}$ does not contain a line that passes through $Q_1$ and a point among  $Q_2,\ldots,Q_8$.

Similarly, a parameter count implies that $\mathcal{Q}$ does not contain secant lines of the curve $\mathscr{C}$,
so that $\mathcal{Q}$ does not contain a line that passes through two points among  $Q_2,\ldots,Q_8$,

Finally, we suppose that $\mathcal{Q}$ contains an irreducible conic $C$ that passes through $Q_1$ and three points among  $Q_2,\ldots,Q_8$.
Let $\rho\colon\mathscr{Q}\dasharrow\mathbb{P}^3$ be a linear projection from the point $Q_1$.
Then $\rho(\mathscr{C})$ is a curve of degree $7$, and the induced map $\mathscr{C}\dasharrow\rho(\mathscr{C})$ is an isomorphism,
because $\mathscr{C}$ is an intersection of quadrics.
Similarly, all points $\rho(Q_2),\ldots,\rho(Q_8)$ are distinct.
Then $\rho(C)$ is a three-secant line of the curve $\rho(\mathscr{C})$.
Note that $\rho(\mathscr{C})$ contains one-parameter family of three-secants  <cit.>.
But $\rho(\mathcal{Q})$ is a general plane in $\mathbb{P}^3$,
which implies that $\rho(\mathcal{Q})$ does not contain three-secant lines of the curve $\rho(\mathscr{C})$ — a contradiction.

Let $\gamma$ be a class in the group $\mathrm{Pic}(T)$ such that $-K_T\cdot \gamma=1$ and $\gamma^2=-1$.
Then the linear system $|\gamma|$ consists of a unique $(-1)$-curve.

Apply the Riemann–Roch theorem, Serre duality and Lemma <ref>.

Let us use Corollary <ref>, to describe infinitely many $(-1)$-curves in the surface $T$.