14. (18 points) Let $x, y, z$ be positive real numbers. Find the minimum value of the function
$$
f(x, y, z)=\frac{(1+2 x)(3 y+4 x)(4 y+3 z)(2 z+1)}{x y z}
$$
14. Given a fixed $y$, we have
$$
\begin{array}{l}
\frac{(1+2 x)(3 y+4 x)}{x}=\frac{8 x^{2}+(6 y+4) x+3 y}{x} \\
=8 x+\frac{3 y}{x}+6 y+4 \\
\geqslant 2 \sqrt{24 y}+6 y+4=(\sqrt{6 y}+2)^{2},
\end{array}
$$

with equality holding if and only if $x=\sqrt{\frac{3 y}{8}}$.
Similarly, $\frac{(4 y+3 z)(2 z+1)}{z}=6 z+\frac{4 y}{z}+8 y+3$
$$
\geqslant(\sqrt{8 y}+\sqrt{3})^{2},
$$

with equality holding if and only if $z=\sqrt{\frac{2 y}{3}}$.
$$
\begin{array}{l}
\text { Therefore, } \frac{(1+2 x)(3 y+4 x)(4 y+3 z)(2 z+1)}{x y z} \\
\geqslant \frac{(\sqrt{6 y}+2)^{2}(\sqrt{8 y}+\sqrt{3})^{2}}{y} \\
\quad=\left(\sqrt{48 y}+\frac{2 \sqrt{3}}{\sqrt{y}}+7 \sqrt{2}\right)^{2} \\
\quad \geqslant(2 \sqrt{\sqrt{48} \times 2 \sqrt{3}}+7 \sqrt{2})^{2} \\
\quad=194+112 \sqrt{3},
\end{array}
$$

with equality holding in the second inequality if and only if $y=\frac{1}{2}$.
Thus, when $x=\frac{\sqrt{3}}{4}, y=\frac{1}{2}, z=\frac{\sqrt{3}}{3}$, $f(x, y, z)$ attains its minimum value of $194+112 \sqrt{3}$.