19. (2006 Zhejiang Province High School Mathematics Party Competition Training Test) Given $a, b, c \in \mathbf{R}^{+}$, and satisfying $\frac{k a b c}{a+b+c} \geqslant(a+b)^{2}$ $+(a+b+4 c)^{2}$, find the minimum value of $k$.
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\text { 19. Solution Since }(a+b)^{2}+(a+b+4 c)^{2}=(a+b)^{2}+[(a+2 c)+(b+2 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} \text {. Therefore, } \frac{(a+b)^{2}+(a+b+4 c)^{2}}{a b c} \cdot(a+b+c) \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}}\right)=100 .
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19. Solution Since $(a+b)^{2}+(a+b+4 c)^{2}=(a+b)^{2}+[(a+2 c)+(b+2 c)]^{2} \geqslant(2 \sqrt{a b})^{2}+(2 \sqrt{2 a c}+2 \sqrt{2 b c})^{2}$ $=4 a b+8 a c+8 b c+16 c \sqrt{a b}$. Therefore, $\frac{(a+b)^{2}+(a+b+4 c)^{2}}{a b c} \cdot(a+b+c) \geqslant 8\left(5 \sqrt[5]{\frac{1}{2 a^{2} b^{2} c}}\right) \cdot\left(5 \sqrt[5]{\frac{a^{2} b^{2} c}{2}}\right)=100$. When $a=b=2 c>0$, the equality holds. Hence, the minimum value of $k$ is 100.
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