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

the minimum value
Let's assume $x \geq y \geq z$.
Case 1: When $256 y^{3} \geq x^{2} z$, we have
$$
\begin{array}{c}
\frac{x^{2}+256 y z}{y^{2}+z^{2}}-\frac{x^{2}}{y^{2}}=\frac{z\left(256 y^{3}-x^{2} z\right)}{\left(y^{2}+z^{2}\right) y^{2}} \geq 0 \\
\frac{y^{2}+256 z x}{z^{2}+x^{2}}-\frac{y^{2}}{x^{2}}=\frac{z\left(256 x^{3}-y^{2} z\right)}{\left(z^{2}+x^{2}\right) x^{2}} \geq 0 \\
\frac{z^{2}+256 x y}{x^{2}+y^{2}}-\frac{256 x y}{x^{2}+y^{2}}=\frac{z^{2}}{x^{2}+y^{2}} \geq 0
\end{array}
$$

Therefore,
$$
\begin{aligned}
f(x, y, z) & \geq \sqrt{\frac{x^{2}}{y^{2}}}+\sqrt{\frac{y^{2}}{x^{2}}}+\sqrt{\frac{256 x y}{x^{2}+y^{2}}}=\frac{x}{y}+\frac{y}{x}+16 \sqrt{\frac{x y}{x^{2}+y^{2}}} \\
& =\frac{x^{2}+y^{2}}{x y}+8 \sqrt{\frac{x y}{x^{2}+y^{2}}}+8 \sqrt{\frac{x y}{x^{2}+y^{2}}} \geq 3 \sqrt[3]{64} \\
& =12
\end{aligned}
$$

Equality holds if and only if $x: y=(2+\sqrt{3}): 1$, and $z=0$, at which point $f(x, y, z)$ reaches 12;
Case 2: When $256 y^{3} < x^{2} z$, we have
$$
\begin{aligned}
f(x, y, z) & \geq \sqrt{\frac{x^{2}+256 y z}{y^{2}+z^{2}}}+0+0>\sqrt{\frac{256 y^{2}+256 y z}{y^{2}+z^{2}}} \\
& =16 \sqrt{\frac{y^{2}+y z}{y^{2}+z^{2}}} \geq 16>12
\end{aligned}
$$

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