2.8 [Pham Kim Hung] Find the greatest constant $k$ such that the following inequality holds for any positive real numbers $a, b, c$
$$\frac{a^{3}+b^{3}+c^{3}}{(a+b)(b+c)(c+a)}+\frac{k(a b+b c+c a)}{(a+b+c)^{2}} \geq \frac{3}{8}+\frac{k}{3}$$
Solution. For $a=b=1+\sqrt{3}$ and $c=1$, we obtain $k \leq \frac{9(3+2 \sqrt{3})}{8}=k_{0}$. We will prove that this is the desired value. Let $k_{0}$ be a constant satisfying the given inequality. Without loss of generality, assume that $p=1$. Then $q \in[0,1]$ and the inequality becomes
$$\frac{3\left(3 r+q^{2}\right)}{-3 r+1-q^{2}}+\frac{k_{0}\left(1-q^{2}\right)}{3} \geq \frac{3}{8}+\frac{k_{0}}{3}$$

It is not difficult to verify that this is an increasing function in terms of $r$. If $2 q \geq 1$, we have
$$V T \geq \frac{3 q^{2}}{1-q^{2}}+\frac{k_{0}\left(1-q^{2}\right)}{3} \geq 1+\frac{k_{0}}{4} \geq \frac{3}{8}+\frac{k_{0}}{3}$$
(since this is an increasing function in terms of $q^{2} \geq \frac{1}{4}$ )
If $2 q \leq 1$, using our theorem, it suffices to prove that
$$\frac{3\left((1+q)^{2}(1-2 q)+9 q^{2}\right)}{-(1+q)^{2}(1-2 q)+9\left(1-q^{2}\right)}+\frac{k_{0}\left(1-q^{2}\right)}{3} \geq \frac{3}{8}+\frac{k_{0}}{3}$$

We have
$$L H S-R H S=\frac{3 q^{2}(3+2 \sqrt{3})(2 \sqrt{3}-1-q)(q-2+\sqrt{3})^{2}}{8(q+1)(q-2)^{2}} \geq 0$$

The inequality is proved, and we conclude that $k_{\max }=k_{0}$.