14. If $a, b, c \in \mathbf{R}^{+}$, and satisfy $\frac{k u b c}{a+b+c} \geqslant(a+b)^{2}+(a+b+4 c)^{2}$. Find the minimum value of $k$.
14. By the mean inequality $(a+b)^{2}+(a+b+4 c)^{2}=(a+b)^{2}+[(a-2 c)+(b+$ $2 c)]^{2} \geqslant(2 \sqrt{a b})^{2}+(2 \sqrt{2 a c}+2 \sqrt{2 b c})^{2}=4 a b+8 a c+8 b c+16 c \sqrt{a b}$, thus
$$
\begin{array}{l}
\frac{(a+b)^{2}+(a+b+4 c)^{2}}{a b c} \cdot(a+b+c) \geqslant \frac{4 a b+8 a c+8 b c+16 c \sqrt{a b}}{a b c} \cdot(a+b+c) \\
=\left(\frac{-4}{c}+\frac{8}{b}+\frac{8}{a}+\frac{16}{\sqrt{a b}}\right)(a+b+c) \\
=8\left(\frac{1}{2 c}+\frac{1}{b}+\frac{1}{a}+\frac{1}{\sqrt{a b}}+\frac{1}{\sqrt{a b}}\right)\left(\frac{a}{2}+\frac{a}{2}+\frac{b}{2}+\frac{b}{2}+c\right) \\
\geqslant 8\left(5 \sqrt[3]{\frac{1}{2 a^{2} b^{2} c}}\right) \cdot\left(5 \sqrt[5]{\frac{a^{2} b^{2} c}{2^{4}}}\right)=100
\end{array}
$$

When $a=b=2 c>0$, the equality holds, hence the maximum value of $k$ is 100.