<s>[INST] Proving $Q=\frac{a^{3}+b^{3}+c^{3}}{(a+b)(b+c)(c+a)}+k\cdot \frac{(ab+bc+ca)}{(a+b+c)^{2}}\geqslant \frac{3}{8}+\frac{k}{3}$ For $a,b,c\geqslant 0.$ Prove: $$\text{Q}=\frac{a^{3}+b^{3}+c^{3}}{(a+b)(b+c)(c+a)}+k\cdot \frac{(ab+bc+ca)}{(a+b+c)^{2}}\geqslant \frac{3}{8}+\frac{k}{3},$$
where $k={\frac {27}{8}}+\frac{9\sqrt{3}}{4}$ is a Root Of $64{k}^{2}-432k-243=0.$
By computer (Maple) I found this inequality is equivalent to$:$ $$\sum \Big[ a+b+ ( 1-\frac{\sqrt {3}}{2} ) c\Big]  \Big[ 2(a+b)- ( 1+\sqrt {3} ) c \Big] ^{2} ( \,a-b \,) ^{2} \geqslant 0$$
But I hope for alternative proof$?$
 [/INST] Another way
Let $p = a + b + c = 3, q = ab+bc+ca, r = abc$.
Since $(a-b)^2(b-c)^2(c-a)^2 \ge 0$ that is $-4p^3r+p^2q^2+18pqr-4q^3-27r^2 \ge 0$,
we have
\begin{align}
r &\ge -\frac{2}{27}p^3 + \frac{1}{3}pq - \frac{2}{27}\sqrt{(p^2 - 3q)^3}\\
&= q - 2 - 2\sqrt{(1-q/3)^3}.\tag{1}
\end{align}
We have to prove that
$$\frac{p^3 - 3pq + 3r}{pq - r} - \frac{3}{8} \ge k\left(\frac{1}{3} - \frac{q}{p^2}\right).$$
From (1), it suffices to prove that
$$\frac{p^3 - 3pq + 3(q - 2 - 2\sqrt{(1-q/3)^3})}{pq - (q - 2 - 2\sqrt{(1-q/3)^3})} - \frac{3}{8} \ge k\left(\frac{1}{3} - \frac{q}{p^2}\right)$$
or
$$\frac{27(1-q/3)(3 - \sqrt{1-q/3})}{8q + 8 + 8\sqrt{(1-q/3)^3}}\ge \frac{k}{3}(1 - q/3).$$
It suffices to prove that
$$\frac{81(3 - \sqrt{1-q/3})}{8q + 8 + 8\sqrt{(1-q/3)^3}}\ge k.$$
With the substitution $q = 3 - \frac{t^2}{3}$ for $0 \le t \le 3$, it suffices to prove that
$$\frac{729(9-t)}{8(t+3)(6-t)^2} \ge k.$$
It is easy to prove that
$$\frac{729(9-t)}{8(t+3)(6-t)^2} - \frac{27}{8}
= \frac{27t^2(9 - t) + 27^2(5 - t)}{8(t+3)(6-t)^2}
> 0$$
and
$$\left(\frac{729(9-t)}{8(t+3)(6-t)^2} - \frac{27}{8}\right)^2 - \frac{243}{16}
= \frac{243(99-6t-t^2)(t^2 - 12t + 9)^2}{64(t+3)^2(6-t)^4} \ge 0$$
which results in
$$\frac{729(9-t)}{8(t+3)(6-t)^2} \ge \frac{27}{8} + \frac{9\sqrt{3}}{4}.$$
(Alternatively, we may use derivatives to find the minimum of $\frac{729(9-t)}{8(t+3)(6-t)^2}$ on $[0, 3]$.)
We are done.
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