Example 10 (1991 National Training Test) Let $x_{1}, x_{2}, x_{3}, x_{4}$ be positive real numbers and $x_{1}+x_{2}+x_{3}+x_{4} = \pi$. Find the minimum value of the expression $\left(2 \sin ^{2} x_{1}+\frac{1}{\sin ^{2} x_{1}}\right)\left(2 \sin ^{2} x_{2}+\frac{1}{\sin ^{2} x_{2}}\right)\left(2 \sin ^{2} x_{3}+\frac{1}{\sin ^{2} x_{3}}\right)\left(2 \sin ^{2} x_{4}+\right.$ $\left.\frac{1}{\sin ^{2} x_{4}}\right)$.
Let $x_{1}+x_{2}$ be a constant. From $\sin x_{1} \cdot \sin x_{2}=\frac{1}{2}\left[\cos \left(x_{1}-x_{2}\right)-\cos \left(x_{1}+x_{2}\right)\right]$, we know that the value of $\sin x_{1} \cdot \sin x_{2}$ increases as $\left|x_{1}-x_{2}\right|$ decreases. Denote the expression under discussion as $f\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$.

If $x_{1}, x_{2}, x_{3}, x_{4}$ are not all equal, without loss of generality, assume $x_{1}>\frac{\pi}{4}>x_{2}$. Let $x_{1}^{\prime}=\frac{\pi}{4}, x_{2}^{\prime}=x_{1}+x_{2}-\frac{\pi}{4}, x_{3}^{\prime}=x_{3}, x_{4}^{\prime}=x_{4}$. Then we have $x_{1}^{\prime}+x_{2}^{\prime}=x_{1}+x_{2},\left|x_{1}^{\prime}-x_{2}^{\prime}\right| < \left|x_{1}-x_{2}\right|$. Therefore, 
\[
\begin{aligned}
f\left(x_{1}, x_{2}, x_{3}, x_{4}\right) & = 2\left(2 \sin ^{2} x_{1} \cdot \sin ^{2} x_{2}+\frac{1}{2 \sin ^{2} x_{1} \cdot \sin ^{2} x_{2}}\right)+2\left(\frac{\sin ^{2} x_{1}}{\sin ^{2} x_{2}}+\frac{\sin ^{2} x_{2}}{\sin ^{2} x_{1}}\right) \\
& = \left(2 \sin ^{2} x_{1}+\frac{1}{\sin ^{2} x_{1}}\right)\left(2 \sin ^{2} x_{2}+\frac{1}{\sin ^{2} x_{2}}\right)
\end{aligned}
\]
Thus, $f\left(x_{1}, x_{2}, x_{3}, x_{4}\right)>f\left(x_{1}^{\prime}, x_{2}^{\prime}, x_{3}^{\prime}, x_{4}^{\prime}\right)$.
If $x_{2}^{\prime}, x_{3}^{\prime}, x_{4}^{\prime}$ are not all equal, we can similarly adjust and prove that when $x_{1}, x_{2}, x_{3}, x_{4}$ are not all equal, we always have $f\left(x_{1}, x_{2}, x_{3}, x_{4}\right)>f\left(\frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{4}\right)$.

Therefore, the minimum value is $f\left(\frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{4}\right)=81$, which is achieved when and only when $x_{1}=x_{2}=x_{3}=x_{4}=\frac{\pi}{4}$.