45a * If positive numbers $a, b, c$ and constant $k$ satisfy the inequality $\frac{k a b c}{a+b+c} \leqslant(a+b)^{2}+(a+b+$ $4 c)^{2}$, find the maximum value of the constant $k$.


Given that $(a+b)^{2}+(a+b+4 c)^{2} \geqslant(2 \sqrt{a b})^{2}+(2 \sqrt{2 a c}+2 \sqrt{2 b c})^{2}=$ $4 a b+8 a c+8 b c+16 c \sqrt{a b}$, therefore
$$
\begin{aligned}
k & =\frac{(a+b)^{2}+(a+b+4 c)^{2}}{a b c} \cdot(a+b+c) \\
& \geqslant \frac{4 a b+8 a c+8 b c+16 c \sqrt{a b}}{a b c}(a+b+c) \\
& =\left(\frac{4}{c}+\frac{8}{b}+\frac{8}{a}+\frac{16}{\sqrt{a b}}\right)(a+b+c) \\
& =8\left(\frac{1}{2 c}+\frac{1}{b}+\frac{1}{a}+\frac{1}{\sqrt{a b}}+\frac{1}{\sqrt{a b}}\right)\left(\frac{a}{2}+\frac{a}{2}+\frac{b}{2}+\frac{b}{2}+c\right) \\
& \geqslant 8\left(5 \sqrt[5]{\frac{1}{2 a^{2} b^{2} c}}\right) \cdot\left(5 \sqrt[5]{\frac{a^{2} b^{2} c^{2}}{2^{4}}}\right)=100,
\end{aligned}
$$

the equality holds if and only if $a=b=2 c$, hence $k_{\text {max }}=100$.