[Example 3.4.6] 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
$$
\begin{array}{l}
\left(2 \sin ^{2} x_{1}+\frac{1}{\sin ^{2} x_{1}}\right) \cdot\left(2 \sin ^{2} x_{2}+\frac{1}{\sin ^{2} x_{2}}\right) . \\
\left(2 \sin ^{2} x_{3}+\frac{1}{\sin ^{2} x_{3}}\right) \cdot\left(2 \sin ^{2} x_{4}+\frac{1}{\sin ^{2} x_{4}}\right)
\end{array}
$$
Let the expression be
$$
f\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=\prod_{i=1}^{4}\left(2 \sin ^{2} x_{i}+\frac{1}{\sin ^{2} x_{i}}\right)
$$

We can guess that when $x_{1}=x_{2}=x_{3}=x_{4}$, $f\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$ reaches its minimum value.
That is, the minimum point is $\left(\frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{4}\right)$.
$$
f\left(x_{1}, x_{2}, x_{3}, x_{4}\right) \geqslant f\left(\frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{4}\right) .
$$

Below, we use the smoothing transformation to prove the above guess.
Keeping $x_{1}+x_{2}+x_{3}+x_{4}=\pi$ unchanged, while transforming $x_{1}, x_{2}, x_{3}, x_{4}$, so that one of the new variables equals the corresponding point of the minimum value.
If $x_{1}, x_{2}, x_{3}, x_{4}$ are not equal, without loss of generality, assume $x_{1}>\frac{\pi}{4}>x_{2}$.

Let $x_{1}^{\prime}=\frac{\pi}{4}, x_{2}^{\prime}=x_{1}+x_{2}-\frac{\pi}{4}, x_{3}^{\prime}=x_{3}, x_{1}^{\prime}=x_{1}$.
At this time, $x_{1}^{\prime}+x_{2}^{\prime}+x_{3}^{\prime}+x_{4}^{\prime}=x_{1}+x_{2}+x_{3}+x_{4}=\frac{\pi}{4}$.
$$
\begin{aligned}
x_{1}^{\prime} x_{2}^{\prime} & =\frac{\pi}{4}\left(x_{1}+x_{2}-\frac{\pi}{4}\right)+x_{1} x_{2}-x_{1} x_{2} \\
& =\left(x_{1}-\frac{\pi}{4}\right)\left(\frac{\pi}{4}-x_{2}\right)+x_{1} x_{2} \\
& >x_{1} x_{2} .
\end{aligned}
$$

Thus, we have
$$
\begin{array}{l}
\left(x_{1}^{\prime}+x_{2}^{\prime}\right)^{2}-\left(x_{1}^{\prime}-x_{2}^{\prime}\right)^{2}>\left(x_{1}+x_{2}\right)^{2}-\left(x_{1}-x_{2}\right)^{2} . \\
\text { That is }\left(x_{1}^{\prime}-x_{2}^{\prime}\right)^{2}x_{2} . \\
\left(2 \sin x_{1}^{2}+\frac{1}{\sin ^{2} x_{1}}\right)\left(2 \sin ^{2} x_{2}+\frac{1}{\sin ^{2} x_{2}}\right) \\
=2\left(2 \sin ^{2} x_{1} \sin ^{2} x_{2}+\frac{1}{2 \sin ^{2} x_{1} \sin ^{2} x_{2}}\right)+2\left(\frac{\sin ^{2} x_{1}}{\sin ^{2} x_{2}}+\frac{\sin ^{2} x_{2}}{\sin ^{2} x_{1}}\right) \text {. } \\
t_{1}=2 \sin ^{2} x_{1} \sin ^{2} x_{2}, t_{2}=\frac{\sin ^{2} x_{2}}{\sin ^{2} x_{1}}, \\
t_{1}^{\prime}=2 \sin ^{2} x_{1}^{\prime} \sin ^{2} x_{2}^{\prime}, t_{2}^{\prime}=\frac{\sin ^{2} x_{2}^{\prime}}{\sin ^{2} x_{1}^{\prime}} \text {. } \\
\end{array}
$$

Also,

Then (1) becomes
$$
\left(2 \sin ^{2} x_{1}+\frac{1}{\sin ^{2} x_{1}}\right)\left(2 \sin ^{2} x_{2}+\frac{1}{\sin ^{2} x_{2}}\right)=2\left(t_{1}+\frac{1}{t_{1}}\right)+2\left(t_{2}+\frac{1}{t_{2}}\right) \text {. }
$$

Also,
$$
\begin{aligned}
& \sin x_{1} \sin x_{2}-\sin x_{1}^{\prime} \sin x_{2}^{\prime} \\
= & \frac{1}{2}\left[\cos \left(x_{1}-x_{2}\right)-\cos \left(x_{1}+x_{2}\right)\right] \\
& -\frac{1}{2}\left[\cos \left(x_{1}^{\prime}-x_{2}^{\prime}\right)-\cos \left(x_{1}^{\prime}+x_{2}^{\prime}\right)\right] \\
= & \frac{1}{2}\left[\cos \left(x_{1}-x_{2}^{\prime}\right)-\cos \left(x_{1}-x_{2}\right)\right]
\end{aligned}
$$
$2\left(t_{1}^{\prime}+\frac{1}{t_{1}^{\prime}}\right)+2\left(t_{2}^{\prime}+\frac{1}{t_{2}^{\prime}}\right) .
$$

Thus, we have
$$
\begin{aligned}
& \left(2 \sin ^{2} x_{1}+\frac{1}{\sin ^{2} x_{1}}\right)\left(2 \sin ^{2} x_{2}+\frac{1}{\sin ^{2} x_{2}}\right) \\
> & \left(2 \sin ^{2} x_{1}^{\prime}+\frac{1}{\sin ^{2} x_{1}^{\prime}}\right)\left(2 \sin ^{2} x_{2}^{\prime}+\frac{1}{\sin ^{2} x_{2}^{\prime}}\right) .
\end{aligned}
$$

Therefore, $\quad f\left(x_{1}, x_{2}, x_{3}, x_{4}\right)>f\left(x_{1}^{\prime}, x_{2}^{\prime}, x_{3}^{\prime}, x_{4}^{\prime}\right)$.
As long as $x_{1}^{\prime}, x_{2}^{\prime}, x_{3}^{\prime}, x_{4}^{\prime}$ are not all equal, we can transform them according to the above method, so that (6) appears,
Therefore, we will have
$$
f\left(x_{1}, x_{2}, x_{3}, x_{4}\right)>\left(\frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{4}\right) .
$$

Thus, the minimum value is
$$
f\left(\frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{4}\right)=81
$$

which is achieved if and only if $x_{1}=x_{2}=x_{3}=x_{1}=\frac{\pi}{4}$.