Solve the following equation:

$$
2^{-\sin ^{2} x}+2^{-\cos ^{2} x}=\sin y+\cos y .
$$
The solution is based on the fact that the minimum of the left side equals the maximum of the right side. The left side is estimated using the arithmetic and geometric mean inequality:

$$
2^{-\sin ^{2} x}+2^{-\cos ^{2} x} \geq 2 \sqrt{2^{-\sin ^{2} x} \cdot 2^{-\cos ^{2} x}}=2 \cdot 2^{-\frac{\sin ^{2} x+\cos ^{2} x}{2}}=\sqrt{2}
$$

and equality holds if $2^{-\sin ^{2} x}=2^{-\cos ^{2} x}$, i.e., $|\sin x|=|\cos x|$, which means $x=\frac{\pi}{4}+k \frac{\pi}{2}$.

For estimating the right side, we use the addition formula of the sine function:

$$
\sin y+\cos y=\sqrt{2}\left(\cos \frac{\pi}{4} \sin y+\sin \frac{\pi}{4} \cos y\right)=\sqrt{2} \sin \left(y+\frac{\pi}{4}\right) \leq \sqrt{2}
$$

equality holds if $y+\frac{\pi}{4}=\frac{\pi}{2}+2 l \pi$, i.e., $y=\frac{\pi}{4}+2 l \pi$.

Thus, the solutions are: $x=\frac{\pi}{4}+k \frac{\pi^{2}}{2}, y=\frac{\pi}{4}+2 l \pi$.

Brezovich Laszló (Szombathely, Nagy Lajos Gymn., III. o)