Let $p, q \in \mathbf{R}_{+}, x \in\left(0, \frac{\pi}{2}\right)$. Find the minimum value of the function $f(x)=\frac{p}{\sqrt{\sin x}}+\frac{q}{\sqrt{\cos x}}$.
Solution: Let $\alpha=\frac{5}{4}, \beta=5$, then, $\frac{1}{\alpha}+\frac{1}{\beta}=1$.
By Hölder's inequality, we have
$$
\begin{array}{l}
p^{\frac{4}{5}}+q^{\frac{4}{5}} \\
=\frac{p^{\frac{4}{5}}}{(\sin x)^{\frac{2}{5}}}(\sin x)^{\frac{2}{5}}+\frac{q^{\frac{4}{5}}}{(\cos x)^{\frac{2}{5}}}(\cos x)^{\frac{2}{5}} \\
\leqslant\left(\frac{p}{\sqrt{\sin x}}+\frac{q}{\sqrt{\cos x}}\right)^{\frac{4}{5}}\left(\sin ^{2} x+\cos ^{2} x\right)^{\frac{1}{5}}, \\
f(x)=\frac{p}{\sqrt{\sin x}}+\frac{q}{\sqrt{\cos x}} \geqslant\left(p^{\frac{4}{5}}+q^{\frac{4}{5}}\right)^{\frac{5}{4}} . \\
\text { The equality holds if and only if } \frac{\frac{p}{\sqrt{\sin x}}}{\frac{q}{\sqrt{\cos x}}}=\frac{\sin ^{2} x}{\cos ^{2} x}, \tan x=\left(\frac{p}{q}\right)^{\frac{2}{5}}
\end{array}
$$

Therefore, the minimum value of $f(x)$ is $\left(p^{\frac{4}{5}}+q^{\frac{4}{5}}\right)^{\frac{5}{4}}$.