Example 7 Given that $a, b$ are positive constants, and $x$ is an acute angle, find the minimum value of the function $y=\frac{a}{\sin ^{n} x}+\frac{b}{\cos ^{n} x}$.
By Hölder's inequality, we have
$$\begin{array}{l}
\left(\frac{a}{\sin ^{n} x}+\frac{b}{\cos ^{n} x}\right)^{\frac{2}{n+2}}\left(\sin ^{2} x+\cos ^{2} x\right)^{\frac{n}{n+2}} \geqslant \\
\left(\frac{a}{\sin ^{n} x}\right)^{\frac{2}{n+2}}\left(\sin ^{2} x\right)^{\frac{n}{n+2}}+\left(\frac{b}{\cos ^{n} x}\right)^{\frac{2}{n+2}}\left(\cos ^{2} x\right)^{\frac{n}{n+2}}= \\
a^{\frac{2}{n+2}}+b^{\frac{2}{n+2}}
\end{array}$$

Thus, 
$$y^{\frac{2}{n+2}} \geqslant a^{\frac{2}{n+2}}+b^{\frac{2}{n+2}}$$

Since $y>0$, we have
$$y \geqslant\left(a^{\frac{2}{n+2}}+b^{\frac{2}{n+2}}\right)^{\frac{n+2}{2}}$$

Equality holds if and only if $\frac{a}{\sin ^{n+2} x}=\frac{b}{\cos ^{n+2} x}$, i.e., $x=\arctan \sqrt[n+2]{\frac{a}{b}}$. Therefore,
$$y_{\min }=\left(a^{\frac{2}{n+2}}+b^{\frac{2}{n+2}}\right)^{\frac{n+2}{2}}$$

When $n=1$, this is a conclusion by Luo Zengru in the 12th issue of *Mathematical Communications* in 1988.