16. (15 points) Let $x, y, z \geqslant 0$, and at most one of them is 0. Find
$$
f(x, y, z)=\sum \sqrt{\frac{x^{2}+256 y z}{y^{2}+z^{2}}}
$$

the minimum value, where $\sum$ denotes the cyclic sum.
16. Let $x \geqslant y \geqslant z$.
(1) When $256 y^{3} \geqslant x^{2} z$, we have
$$
\begin{array}{l}
\frac{x^{2}+256 y z}{y^{2}+z^{2}}-\frac{x^{2}}{y^{2}}=\frac{z\left(256 y^{3}-x^{2} z\right)}{\left(y^{2}+z^{2}\right) y^{2}} \geqslant 0, \\
\frac{y^{2}+256 z x}{z^{2}+x^{2}}-\frac{y^{2}}{x^{2}}=\frac{z\left(256 x^{3}-y^{2} z\right)}{\left(z^{2}+x^{2}\right) x^{2}} \geqslant 0, \\
\frac{z^{2}+256 x y}{x^{2}+y^{2}}-\frac{256 x y}{x^{2}+y^{2}}=\frac{z^{2}}{x^{2}+y^{2}} \geqslant 0,
\end{array}
$$

Thus,
$$
\begin{array}{l}
f(x, y, z) \\
\geqslant \sqrt{\frac{x^{2}}{y^{2}}}+\sqrt{\frac{y^{2}}{x^{2}}}+\sqrt{\frac{256 x y}{x^{2}+y^{2}}} \\
=\frac{x^{2}+y^{2}}{x y}+8 \sqrt{\frac{x y}{x^{2}+y^{2}}}+8 \sqrt{\frac{x y}{x^{2}+y^{2}}} \\
\geqslant 3 \sqrt[3]{64}=12,
\end{array}
$$

with equality if and only if $x=(2+\sqrt{3}) y$ and $z=0$, in which case $f(x, y, z)$ achieves 12.
(2) When $256 y^{3} < x^{2} z$, we have
$$
\begin{array}{l}
f(x, y, z) \\
\geqslant \sqrt{\frac{x^{2}+256 y z}{y^{2}+z^{2}}}+\sqrt{\frac{y^{2}+256 z x}{z^{2}+x^{2}}}+\sqrt{\frac{z^{2}+256 x y}{x^{2}+y^{2}}} \\
\geqslant \sqrt{\frac{x^{2}}{y^{2}}}+\sqrt{\frac{y^{2}}{x^{2}}}+\sqrt{\frac{256 x y}{x^{2}+y^{2}}} \\
\geqslant \sqrt{\frac{256 y^{2}+256 z^{2}}{y^{2}+z^{2}}}+0+0 \\
=16>12 .
\end{array}
$$

In summary, $f(x, y, z)_{\min }=12$.