$9 \cdot 60$ Let $x_{1}, x_{2}, x_{3}, x_{4}$ all be positive numbers, and $x_{1}+x_{2}+x_{3}+x_{4}=\pi$, find the minimum value of the expression
$$\left(2 \sin ^{2} x_{1}+\frac{1}{\sin ^{2} x_{1}}\right)\left(2 \sin ^{2} x_{2}+\frac{1}{\sin ^{2} x_{2}}\right)\left(2 \sin ^{2} x_{3}+\frac{1}{\sin ^{2} x_{3}}\right)\left(2 \sin ^{2} x_{4}+\frac{1}{\sin ^{2} x_{4}}\right)$$
[Solution] By the AM-GM inequality, we have
$$2 \sin ^{2} x_{i}+\frac{1}{\sin ^{2} x_{i}}=2 \sin ^{2} x_{i}+\frac{1}{2 \sin ^{2} x_{i}}+\frac{1}{2 \sin ^{2} x_{i}} \geqslant 3 \sqrt[3]{\frac{1}{2 \sin ^{2} x_{i}}}$$

Therefore, $\prod_{i=1}^{4}\left(2 \sin ^{2} x_{i}+\frac{1}{\sin ^{2} x_{i}}\right) \geqslant 81\left(4 \sin x_{1} \sin x_{2} \sin x_{3} \sin x_{4}\right)^{-\frac{2}{3}}$.
By the concavity of the function $f(x)=\ln \sin x$ in $(0, \pi)$, we have
$$\sin x_{1} \sin x_{2} \sin x_{3} \sin x_{4} \leqslant \sin ^{4} \frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}=\frac{1}{4},$$

Thus, $\prod_{i=1}^{4}\left(2 \sin ^{2} x_{i}+\frac{1}{\sin ^{2} x_{i}}\right) \geqslant 81$.
When $x_{1}=x_{2}=x_{3}=x_{4}=\frac{\pi}{4}$, $\prod_{i=1}^{4}\left(2 \sin ^{2} x_{i}+\frac{1}{\sin ^{2} x_{i}}\right)=81$, hence 81 is the minimum value sought.