Example 14 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 $\left(2 \sin ^{2} x_{1}+\right.$ $\left.\frac{1}{\sin ^{2} x_{1}}\right)\left(2 \sin ^{2} x_{2}+\frac{1}{\sin ^{2} x_{2}}\right)\left(2 \sin ^{2} x_{3}+\frac{1}{\sin ^{2} x_{3}}\right)\left(2 \sin ^{2} x_{4}+\frac{1}{\sin ^{2} x_{4}}\right)$.
By the AM-GM inequality, we have:
$2 \sin ^{2} x_{i}+\frac{1}{\sin ^{2} x_{i}}=2 \sin ^{2} x_{i}+\frac{1}{2 \sin ^{2} x_{i}}+\frac{1}{2 \sin ^{2} x_{i}} \geqslant 3 \sqrt[3]{\frac{1}{2 \sin ^{2} x_{i}}}, i=1,2,3,4$.
Thus, $\prod_{i=1}^{4}\left(2 \sin ^{2} x_{i}+\frac{1}{\sin ^{2} x_{i}}\right) \geqslant 81\left(4 \sin x_{1} \sin x_{2} \sin x_{3} \sin x_{4}\right)^{-\frac{2}{3}}$.
Since $f(x)=\ln \sin x$ is a concave function for $x \in(0, \pi)$.
Hence, $\sin x_{1} \cdot \sin x_{2} \cdot \sin x_{3} \cdot \sin x_{4} \leqslant \sin ^{4} \frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}=\frac{1}{4}$.
Therefore, $\prod_{i=1}^{4}\left(2 \sin ^{2} x_{i}+\frac{1}{\sin ^{2} x_{i}}\right) \geqslant 81$.
Moreover, when $x_{1}=x_{2}=x_{3}=x_{4}=\frac{\pi}{4}$, $\prod_{i=1}^{4}\left(2 \sin ^{2} x_{i}+\frac{1}{\sin ^{2} x_{i}}\right)=81$.
Thus, the minimum value of the given expression is 81.