<s>[INST] Inequality with $a+b+c=1$ Let $a,b,c\in\mathbb{R^+}$ such that $a+b+c=1$. Prove that $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+3(ab+bc+ca)\geq\frac{11}{2}.$$
I am trying to resolve this problem but actually i found some issues:
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=3+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{9}{2}$$
I only use Nesbitt's inequality. Then we only need that $3(ab+bc+ca)\geq 1$, but it is not correct, because $3(ab+bc+ca)\leq 1$.
Maybe some ideas will be more that grateful.
 [/INST] Variational Approach
Let $x=a+b$, $y=b+c$, and $z=c+a$. Then $x+y+z=2$ and we are minimizing
$$
\frac1x+\frac1y+\frac1z+3\left(1-\frac{x^2+y^2+z^2}2\right)\tag{1}
$$
Furthermore, since $a+b+c=1$ and $a,b,c\ge0$, we have that $0\le x,y,z\le1$.
Since $x+y+z=2$, we get the restriction
$$
\delta x+\delta y+\delta z=0\tag{2}
$$
Taking the variation of $(1)$ gives
$$
\left(\frac1{x^2}+3x\right)\delta x+\left(\frac1{y^2}+3y\right)\delta y+\left(\frac1{z^2}+3z\right)\delta z=0\tag{3}
$$
Thus, to minimize $(1)$, we need to have $(3)$ whenever we have $(2)$. This means that
$$
\frac1{x^2}+3x=\frac1{y^2}+3y=\frac1{z^2}+3z\tag{4}
$$

Case $\boldsymbol{1}$: If $x\ne y$ and $y\ne z$ and $z\ne x$, then $(4)$ implies
$$
\frac{x+y}{x^2y^2}=\frac{y+z}{y^2z^2}=\frac{z+x}{z^2x^2}=3\tag{5}
$$
which implies that $x=y=z$.

Case $\boldsymbol{2}$: If $x=y\ne z$, then $2x+z=2$ and
$$
\frac{z+x}{z^2x^2}=3\implies12x^4-24x^3+12x^2+x-2=0\tag{6}
$$
Sturm's Theorem shows that there are no solutions to $(6)$ in $[0,1]$.

Therefore, $(1)$ is minimized when $x=y=z=\frac23$ and thus, $a=b=c=\frac13$. This means that
$$
\frac1{a+b}+\frac1{b+c}+\frac1{c+a}+3(ab+bc+ca)\ge\frac{11}2\tag{7}
$$
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