10. (20 points) Given the function $f(x)=\frac{1}{x}-x$. If for any $x \in(0, k)$, we have $f(x) \cdot f(k-x) \geqslant \left(\frac{k}{2}-\frac{2}{k}\right)^{2}$ always holds. Find the range of positive real number $k$.
10. Let $y=k-x \in(0, k)$, then
$$
\begin{aligned}
f(x) f(k-x) & =f(x) f(y)=\left(\frac{1}{x}-x\right)\left(\frac{1}{y}-y\right) \\
& =\frac{x^{2} y^{2}-\left(x^{2}+y^{2}\right)+1}{x y} \\
& =x y+\frac{1-k^{2}}{x y}+2 .
\end{aligned}
$$

Let $x y=t$, then $t \in\left(0, \frac{k^{2}}{4}\right]$. Let $g(t)=t+\frac{1-k^{2}}{t}+2, t \in\left(0, \frac{k^{2}}{4}\right]$.
(1) When $1-k^{2} \leqslant 0$, $g(t)$ is monotonically increasing on $\left(0, \frac{k^{2}}{4}\right]$, then
$$
g(t)_{\max }=g\left(\frac{k^{2}}{4}\right)=\left(\frac{k}{2}-\frac{2}{k}\right)^{2}
$$

This does not meet the condition, discard.
(2) When $1-k^{2}>0$, $g(t)$ is monotonically decreasing on $\left(0, \sqrt{1-k^{2}}\right]$, and monotonically increasing on $\left[\sqrt{1-k^{2}}, \frac{k^{2}}{4}\right]$. Therefore, $g(t)_{\min}=g\left(\sqrt{1-k^{2}}\right)=2\sqrt{1-k^{2}}+2$, and $g\left(\frac{k^{2}}{4}\right)=\left(\frac{k}{2}-\frac{2}{k}\right)^{2}$. For $0<k<2$, $\sqrt{1-k^{2}}<\frac{k^{2}}{4}$, and $g\left(\frac{k^{2}}{4}\right)>g\left(\sqrt{1-k^{2}}\right)$, i.e., $\left(\frac{k}{2}-\frac{2}{k}\right)^{2}>g\left(\sqrt{1-k^{2}}\right)$, which does not meet the condition, discard.
In summary, the range of real number $k$ is $\{k \mid 0<k \leqslant \sqrt{\sqrt{5}-2}\}$.