Let $n$ be a positive integer. Solve the equation

$$
\frac{1}{\sin ^{2 n} x}+\frac{1}{\cos ^{2 n} x}=2^{n+1}
$$
Solution. The denominator cannot be zero, so $\sin ^{2 n} x \neq 0$, and $\cos ^{2 n} x \neq 0$. Therefore, $x \neq k \cdot \frac{\pi}{2}, k \in \mathbb{Z}$.

To solve this, we use the application of notable means. Divide the equation by 2, then take the reciprocal of both sides:

$$
\frac{2}{\frac{1}{\sin ^{2 n} x}+\frac{1}{\cos ^{2 n} x}}=\frac{1}{2^{n}}
$$

It is immediately clear that the left side is in the form $\frac{2}{\frac{1}{a}+\frac{1}{b}}$, which is the harmonic mean of the positive numbers $\sin ^{2 n} x$ and $\cos ^{2 n} x$. Now we use the known relations among means:

$$
\frac{2}{\frac{1}{a}+\frac{1}{b}} \leq \sqrt{a b} \leq \frac{a+b}{2}
$$

$$
\begin{aligned}
\left(\frac{1}{2}\right)^{n} & =\frac{2}{\frac{1}{\sin ^{2 n} x}+\frac{1}{\cos ^{2 n} x}} \leq \sqrt{\sin ^{2 n} x \cdot \cos ^{2 n} x}=\sqrt{\left(\sin ^{2} x \cdot \cos ^{2} x\right)^{n}}= \\
& =\left(\sqrt{\sin ^{2} x \cdot \cos ^{2} x}\right)^{n} \leq\left(\frac{\sin ^{2} x+\cos ^{2} x}{2}\right)^{n}=\left(\frac{1}{2}\right)^{n}
\end{aligned}
$$

Equality in the means holds if and only if $\sin ^{2} x=\cos ^{2} x$, which means $x=(2 k+1) \frac{\pi}{4}$, where $k \in \mathbb{Z}$.

Remark. Using the means, we also showed that the left side of the original equation is always greater than or equal to the right side.