2. Let $x+y=k, x, y \in \mathbf{R}^{+}$. Try to find the maximum value of $k$ such that the inequality $\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right) \geqslant\left(\frac{k}{2}+\frac{2}{k}\right)^{2}$ always holds.
(Notice that when $x=y$, the above equation obviously holds. Therefore, without loss of generality, assume $x>y$, and use the mean substitution: let $m=\frac{k}{2}, x=m+t, y=m-t, 0<t<m$, rewrite the original inequality as
$$
\left(m+t+\frac{1}{m+t}\right)\left(m-t+\frac{1}{m-t}\right) \geqslant\left(m+\frac{1}{m}\right)^{2} \text {. }
$$

That is, $r^{2} \geqslant \frac{m^{4}-4 m^{2}-1}{m^{2}}$.
Answer: $k_{\text {pont }}=2 \sqrt{2+\sqrt{5}}$.)