40. Find the best constant $k$ such that the following inequality holds for all positive real numbers $a, b, c$.
$$\frac{(a+b)(b+c)(c+a)}{a b c}+\frac{k(a b+b c+c a)}{a^{2}+b^{2}+c^{2}} \geq 8+k \quad(\text{Pham Kim Hung})$$
Solution: We have
$$\begin{array}{l}
\frac{(a+b)(b+c)(c+a)}{a b c}-8=\frac{c(a-b)^{2}+a(b-c)^{2}+b(c-a)^{2}}{a b c} \\
1-\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}}=\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{2\left(a^{2}+b^{2}+c^{2}\right)}
\end{array}$$

So we only need to find a positive constant $k$ satisfying the condition
$$\sum_{s y m}(a-b)^{2}\left[\frac{2\left(a^{2}+b^{2}+c^{2}\right)}{a b}-k\right] \geq 0 \Leftrightarrow \sum_{s y m}(a-b)^{2} S_{c} \geq 0$$

where $S_{a}, S_{b}, S_{c}$ are defined by the following expressions
$$\begin{array}{l}
S_{a}=2 a\left(a^{2}+b^{2}+c^{2}\right)-k a b c \\
S_{b}=2 b\left(a^{2}+b^{2}+c^{2}\right)-k a b c \\
S_{c}=2 c\left(a^{2}+b^{2}+c^{2}\right)-k a b c
\end{array}$$
(i) Necessary condition. If $b=c$, we have $S_{b}=S_{c}$; therefore, if (*) holds, we must have
$$S_{b} \geq 0 \Leftrightarrow 2\left(a^{2}+2 b^{2}\right) \geq k a b$$

By the AM-GM inequality, we find the best positive constant $k$ is $4 \sqrt{2}$
(ii) Sufficient condition: For $k \leq 4 \sqrt{2}$, we will prove that the inequality always holds. Without loss of generality, we assume $a \geq b \geq c$, then $S_{a} \geq S_{b} \geq S_{c}$. Of course, $S_{a}=2 a\left(a^{2}+b^{2}+c^{2}\right)-k a b c \geq 0$. Let $x=\sqrt{b c}$, then
$$S_{b}+S_{c}=2(b+c)\left(a^{2}+b^{2}+c^{2}\right)-2 k a b c \geq 4 x\left(a^{2}+2 x^{2}\right)-2 k a x^{2}=4 x(a-\sqrt{2} x)^{2} \geq 0$$

Thus, we have
$$\sum_{c y c} S_{a}(b-c)^{2} \geq\left(S_{b}+S_{c}\right)(a-b)^{2} \geq 0$$

Conclusion: The best positive constant $k$ value is $4 \sqrt{2}$. If $k=4 \sqrt{2}$, the equality holds when $a=b=c$ or $a=\sqrt{2} b=\sqrt{2} c$ and their permutations. If $k<4 \sqrt{2}$, the equality holds when $a=b=c$.