Example $\mathbf{1}$ Given $\theta_{1}, \theta_{2}, \theta_{3}, \theta_{4} \in \mathbf{R}^{+}$ and $\theta_{1}+\theta_{2}+\theta_{3}+\theta_{4}=\pi$, find the minimum value of $\left(2 \sin ^{2} \theta_{1}+\frac{1}{\sin ^{2} \theta_{1}}\right)\left(2 \sin ^{2} \theta_{2}+\frac{1}{\sin ^{2} \theta_{2}}\right)\left(2 \sin ^{2} \theta_{3}+\frac{1}{\sin ^{2} \theta_{3}}\right)\left(2 \sin ^{2} \theta_{1}+\frac{1}{\sin ^{2} \theta_{1}}\right)$.
Analyzing the symmetry conjecture that $\theta_{1}=\theta_{2}=\theta_{3}=\theta_{4}=\frac{\pi}{4}$, we obtain the minimum value, and then use the AM-GM inequality.
Solution: 
\[ 2 \sin ^{2} \theta_{1}+\frac{1}{2 \sin ^{2} \theta_{1}}+\frac{1}{2 \sin ^{2} \theta_{1}} \geqslant 3 \sqrt[3]{2 \sin ^{2} \theta_{1} \cdot\left(\frac{1}{2 \sin ^{2} \theta_{1}}\right)^{2}}=3 \sqrt[3]{\frac{1}{2 \sin ^{2} \theta_{1}}} \]
Thus,
\[ 2 \sin ^{2} \theta_{1}+\frac{1}{\sin ^{2} \theta_{1}} \geqslant 3 \sqrt[3]{\frac{1}{2 \sin ^{2} \theta_{1}}} \]
Similarly, we have:
\[ 2 \sin ^{2} \theta_{i}+\frac{1}{\sin ^{2} \theta_{i}} \geqslant 3 \sqrt[3]{\frac{1}{2 \sin ^{2} \theta_{i}}} \quad (i=1,2,3,4) \]
Equality holds when $2 \sin ^{2} \theta_{i}=\frac{1}{2 \sin ^{2} \theta_{i}}$, i.e., $\sin \theta_{i}=\frac{\sqrt{2}}{2}, \theta_{i}=\frac{\pi}{4}$.
Therefore, the original expression is:
\[ \geqslant 3^{4} \sqrt[3]{\frac{1}{2^{4} \sin ^{2} \theta_{1} \sin ^{2} \theta_{2} \sin ^{2} \theta_{3} \sin ^{2} \theta_{4}}} \]
\[
\text{And } \begin{aligned}
\sin \theta_{1} \cdot \sin \theta_{2} \cdot \sin \theta_{3} \cdot \sin \theta_{1} & \leqslant\left(\frac{\sin \theta_{1}+\sin \theta_{2}+\sin \theta_{3}+\sin \theta_{4}}{4}\right)^{4} \\
& =\left(\frac{2 \sin \frac{\theta_{1}+\theta_{2}}{2} \cos \frac{\theta_{1}-\theta_{2}}{2}+2 \sin \frac{\theta_{3}+\theta_{4}}{2} \cos \frac{\theta_{3}-\theta_{4}}{2}}{4}\right)^{4} \\
& \leqslant\left(\frac{2 \sin \frac{\theta_{1}+\theta_{2}}{2}+2 \sin \frac{\theta_{3}+\theta_{4}}{2}}{4}\right)^{4} \\
& \leqslant\left(\frac{4 \sin \frac{\theta_{1}+\theta_{2}+\theta_{3}+\theta_{4}}{4}}{4}\right. \\
& =\sin ^{4} \frac{\pi}{4}=\frac{1}{4}
\end{aligned}
\]
Equality holds when $\theta_{1}=\theta_{2}=\theta_{3}=\theta_{1}=\frac{\pi}{4}$.
Therefore, the original expression is:
\[ \geqslant 3^{4} \sqrt[3]{\frac{1}{2^{4}\left(\frac{1}{4}\right)^{2}}}=3^{4} \]
Equality holds when $\theta_{1}=\theta_{2}=\theta_{3}=\theta_{4}$.