Question 65, Positive real numbers $\mathrm{x}>\mathrm{y}$, find $\max \left\{\mathrm{x}^{2}, \frac{25}{\mathrm{y}(\mathrm{x}-\mathrm{y})}\right\}$'s minimum value.
Question 65, Solution: According to the AM-GM inequality, we have:
$$
\begin{array}{l}
2 \times \max \left\{x^{2}, \frac{25}{y(x-y)}\right\} \\
\geq x^{2}+\frac{25}{y(x-y)} \\
=[y+(x-y)]^{2}+\frac{25}{y(x-y)} \\
\geq 4 y(x-y)+\frac{25}{y(x-y)} \\
\geq 20
\end{array}
$$

Therefore, $\max \left\{x^{2}, \frac{25}{y(x-y)}\right\} \geq 10$. When $x=\sqrt{10}, y=\frac{\sqrt{10}}{2}$, $\max \left\{x^{2}, \frac{25}{y(x-y)}\right\}=10$.