16. Given positive real numbers $x, y, z$ satisfying
$$
(x+y+z) x y z=4 \text {. }
$$

Find the minimum value of $(x+y)^{2}+(y+z)^{2}+(z+x)^{2}$.
16. Let $x+y=a, y+z=b, z+x=c$. Then $a$, $b$, $c$ are the side lengths of $\triangle ABC$.
From $(x+y+z)xyz=4$
$\Rightarrow \sqrt{(x+y+z)xyz}=2$
$\Rightarrow \sqrt{p(p-a)(p-b)(p-c)}=2$,

where $p=\frac{1}{2}(a+b+c)$.
Establish a Cartesian coordinate system, set $A(0,0), B(c, 0)$, $C(m, n)$.
Then $S_{\triangle ABC}=\frac{1}{2} c|n|=2$, and
$(x+y)^{2}+(y+z)^{2}+(z+x)^{2}$
$=a^{2}+b^{2}+c^{2}=(c-m)^{2}+n^{2}+m^{2}+n^{2}+c^{2}$
$=2m^{2}-2cm+2n^{2}+2c^{2}$
$=2\left(m-\frac{c}{2}\right)^{2}+2n^{2}+\frac{3}{2}c^{2}$
$\geqslant 2n^{2}+\frac{3}{2}c^{2}$ (when and only when $m=\frac{c}{2}$ the equality holds)
$\geqslant 2\sqrt{3}c|n|$
$=8\sqrt{3}$ (when and only when $m=\frac{c}{2},|n|=\frac{\sqrt{3}}{2}c$ the equality holds).
Therefore, the minimum value of $(x+y)^{2}+(y+z)^{2}+(z+x)^{2}$ is $8\sqrt{3}$.