4.5.5 ** If for all positive integers $m, n$ the inequality
$$
\left(\frac{1}{m+n}\right)^{2} \leqslant k\left(\frac{1}{m^{2}}+\frac{1}{n^{2}}\right)
$$

holds, find the minimum value of the real number $k$.
Given that $\frac{1}{(m+n)^{2}} \leqslant k\left(\frac{1}{m^{2}}+\frac{1}{n^{2}}\right) \Leftrightarrow k \geqslant \frac{1}{(m+n)^{2}\left(\frac{1}{m^{2}}+\frac{1}{n^{2}}\right)}$, and because
$$
\begin{aligned}
(m+n)^{2}\left(\frac{1}{m^{2}}+\frac{1}{n^{2}}\right) & =\left(1+\frac{n}{m}\right)^{2}+\left(1+\frac{m}{n}\right)^{2} \\
& \geqslant 2 \sqrt{\left(1+\frac{n}{m}\right)^{2}\left(1+\frac{m}{n}\right)^{2}}=2\left(2+\frac{n}{m}+\frac{m}{n}\right) \\
& \geqslant 2\left(2+2 \sqrt{\frac{n}{m} \cdot \frac{m}{n}}\right)-8 \text { (equality holds if and only if } m=n \text {), }
\end{aligned}
$$

thus, $\frac{1}{(m+n)^{2}\left(\frac{1}{m^{2}}+\frac{1}{n^{2}}\right)} \leqslant \frac{1}{8}$. Therefore, $k_{\min }=\frac{1}{8}$.