16. Let the positive even function $f(x)$ satisfy $f(1)=2$, and when $x y \neq 0$, $f\left(\sqrt{x^{2}+y^{2}}\right) \leqslant \frac{f(x) f(y)}{f(x)+f(y)}$, then $f(5)$ $\qquad$ $\frac{2}{25}$ (fill in one of $>,<, \geqslant,=, \leqslant$ ).
16. $\leqslant$ From the condition, we have $\frac{1}{f\left(\sqrt{x^{2}+y^{2}}\right)} \geqslant \frac{1}{f(x)}+\frac{1}{f(y)}$.

Since $f(-x)=f(x)$, we only need to consider the case where $x, y \in \mathbf{R}^{+}$.
Let
then
$$
\begin{array}{c}
x=\sqrt{u}, y=\sqrt{v}, \\
\frac{1}{f(\sqrt{u+v})} \geqslant \frac{1}{f(\sqrt{u})}+\frac{1}{f(\sqrt{v})} . \\
g(t)=\left\{\begin{array}{l}
-\frac{1}{f(\sqrt{t})}, f(\sqrt{t}) \neq 0 ; \\
0, f(\sqrt{t})=0,
\end{array}\right.
\end{array}
$$

Let again

then it becomes
$$
g(u+v) \geqslant g(u)+g(v) .
$$

From equation (1), we know $g(2 u)=g(u+u) \geqslant 2 g(u)$,
$$
g(3 u)=g(u+2 u) \geqslant g(u)+2 g(u)=3 g(u) .
$$

By induction, we get $g(k u) \geqslant k g(u)$.
Let $k=25, u=1$, then $g(25) \geqslant 25 g(1)$, i.e., $\frac{1}{f(\sqrt{25})} \geqslant \frac{25}{f(\sqrt{1})} \Rightarrow f(5) \leqslant \frac{2}{25}$.