16. If $a>0, b>0$, then
$$
\min \left\{\max \left\{a, b, \frac{1}{a^{2}}+\frac{1}{b^{2}}\right\}\right\}=
$$
$\qquad$
16. $\sqrt[3]{2}$.
$$
\begin{array}{l}
\text { Let } \max \left\{a, b, \frac{1}{a^{2}}+\frac{1}{b^{2}}\right\}=m \text {. Then } \\
a \leqslant m, b \leqslant m, \frac{1}{a^{2}}+\frac{1}{b^{2}} \leqslant m \\
\Rightarrow m \geqslant \frac{2}{m^{2}} \Rightarrow m \geqslant \sqrt[3]{2} \\
\Rightarrow \min \left\{\max \left\{a, b, \frac{1}{a^{2}}+\frac{1}{b^{2}}\right\}\right\}=\sqrt[3]{2} .
\end{array}
$$