Let
$$
D=\left\{(x_{1}, x_{2}) \mid x_{1}>0, x_{2}>0, x_{1}+x_{2}=k\right\} \text {. }
$$
(1) Find the range of real numbers $k$ such that the inequality
$$
\left(\frac{1}{x_{1}}-x_{1}\right)\left(\frac{1}{x_{2}}-x_{2}\right) \geqslant\left(\frac{2}{k}-\frac{k}{2}\right)^{2}
$$

holds for all $\left(x_{1}, x_{2}\right) \in D$;
(2) Find the range of real numbers $k$ such that the inequality
$$
\left(\frac{1}{x_{1}}-x_{1}\right)\left(\frac{1}{x_{2}}-x_{2}\right) \leqslant\left(\frac{2}{k}-\frac{k}{2}\right)^{2}
$$

holds for all $\left(x_{1}, x_{2}\right) \in D$.
Solution: Note that $\left(x_{1}-x_{2}\right)^{2}=k^{2}-4 x_{1} x_{2}$, then
$$
\begin{aligned}
P & =\left(\frac{1}{x_{1}}-x_{1}\right)\left(\frac{1}{x_{2}}-x_{2}\right)-\left(\frac{2}{k}-\frac{k}{2}\right)^{2} \\
& =\frac{1}{x_{1} x_{2}}+x_{1} x_{2}-\frac{x_{2}}{x_{1}}-\frac{x_{1}}{x_{2}}-\frac{4}{k^{2}}-\frac{k^{2}}{4}+2 \\
& =\frac{1}{x_{1} x_{2}}-\frac{4}{k^{2}}-\left(\frac{k^{2}}{4}-x_{1} x_{2}\right)-\left(\frac{x_{2}}{x_{1}}+\frac{x_{1}}{x_{2}}-2\right) \\
& =\frac{k^{2}-4 x_{1} x_{2}}{k^{2} x_{1} x_{2}}-\frac{k^{2}-4 x_{1} x_{2}}{4}-\frac{\left(x_{1}-x_{2}\right)^{2}}{x_{1} x_{2}} \\
& =\frac{\left(k^{2}-4 x_{1} x_{2}\right)\left(4-k^{2} x_{1} x_{2}\right)-4 k^{2}\left(x_{1}-x_{2}\right)^{2}}{4 k^{2} x_{1} x_{2}} \\
& =\frac{\left(x_{1}-x_{2}\right)^{2}\left(4-k^{2} x_{1} x_{2}-4 k^{2}\right)}{4 k^{2} x_{1} x_{2}}
\end{aligned}
$$
(1) Substituting $x_{1} x_{2}=\frac{k^{2}-\left(x_{1}-x_{2}\right)^{2}}{4}$ into the above expression, we have
$$
P=\frac{\left(x_{1}-x_{2}\right)^{2}\left[16-h^{4}-16 k^{2}+k^{2}\left(x_{1}-x_{2}\right)^{2}\right]}{16 k^{2} x_{1} x_{2}}.
$$

Thus, inequality (1) is equivalent to
$$
\frac{\left(x_{1}-x_{2}\right)^{2}\left[16-k^{4}-16 k^{2}+k^{2}\left(x_{1}-x_{2}\right)^{2}\right]}{16 k^{2} x_{1} x_{2}} \geqslant 0.
$$
Clearly, equation (3) holds for any $\left(x_{1}, x_{2}\right) \in D$, i.e.,
$$
16-k^{4}-16 k^{2}+k^{2}\left(x_{1}-x_{2}\right)^{2} \geqslant 0
$$

The necessary and sufficient condition for the above inequality to hold for any $\left(x_{1}, x_{2}\right) \in D$ is
$$
k^{4}+16 k^{2}-16 \leqslant \min _{1,1, x_{2} \in D}\left[k^{2}\left(x_{1}-x_{2}\right)^{2}\right]=0.
$$

Noting that $k>0$, solving yields
$$
00$, solving yields $k \geqslant 1$.
Therefore, the range of $k$ that makes inequality (2) hold for any $\left(x_{1}, x_{2}\right) \in D$ is $[1,+\infty)$.
(Jiang Mingbin, Peng'an High School, Sichuan, 637851)