Problem 40. Find the best positive real constant $k$ such that the following inequality holds for all positive real numbers $a, b$ and $c$
$$\frac{(a+b)(b+c)(c+a)}{a b c}+\frac{k(a b+b c+c a)}{a^{2}+b^{2}+c^{2}} \geq 8+k$$
Solution. We clearly have
$$\begin{aligned}
\frac{(a+b)(b+c)(c+a)}{a b c}-8 & =\frac{c(a-b)^{2}+a(b-c)^{2}+b(c-a)^{2}}{a b c} \\
1-\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}} & =\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{2\left(a^{2}+b^{2}+c^{2}\right)}
\end{aligned}$$

So we need to find a positive number $k$ satisfying the condition
$$\sum_{\text {sym }}(a-b)^{2}\left(\frac{2\left(a^{2}+b^{2}+c^{2}\right)}{a b}-k\right) \geq 0 \Leftrightarrow \sum_{\text {sym }}(a-b)^{2} S_{c} \geq 0(\star)$$
where $S_{a}, S_{b}, S_{c}$ are defined as
$$\begin{array}{l}
S_{a}=2 a\left(a^{2}+b^{2}+c^{2}\right)-k a b c \\
S_{b}=2 b\left(a^{2}+b^{2}+c^{2}\right)-k a b c \\
S_{c}=2 c\left(a^{2}+b^{2}+c^{2}\right)-k a b c
\end{array}$$
(i). Necessary condition: If $b=c$, we have $S_{b}=S_{c}$; so if $(\star)$ is true, we must have
$$S_{b} \geq 0 \Leftrightarrow 2\left(a^{2}+2 b^{2}\right) \geq k a b$$

By AM-GM inequality, we find that the best value of $k$ is $4 \sqrt{2}$.
(ii). Sufficient condition: For $k \leq 4 \sqrt{2}$, we will prove that the inequality is always true. WLOG, we may assume that $a \geq b \geq c$. Then $S_{a} \geq S_{b} \geq S_{c}$. Certainly, $S_{a}=2 a\left(a^{2}+b^{2}+c^{2}\right)-k a b c \geq 0$. Let $x=\sqrt{b c}$, then
$$\begin{array}{c}
S_{b}+S_{c}=2(b+c)\left(a^{2}+b^{2}+c^{2}\right)-2 k a b c \geq \\
\geq 4 x\left(a^{2}+2 x^{2}\right)-2 k a x^{2}=4 x(a-\sqrt{2} x)^{2} \geq 0
\end{array}$$

We conclude that
$$\sum_{c y c} S_{a}(b-c)^{2} \geq\left(S_{b}+S_{c}\right)(a-b)^{2} \geq 0$$

Conclusion: The best value of $k$ is $4 \sqrt{2}$. If $k=4 \sqrt{2}$, equality holds for $a=b=c$ or $a=\sqrt{2} b=\sqrt{2} c$ up to permutation. If $k<4 \sqrt{2}$, the equality holds only for $a=b=c$.