Let $S$ be the square $0\le x\le\frac\pi2$, $0\le y\le\frac\pi2$ in the coordinate plane. What is the area of the region of $S$ where $\cos^2x+\cos^2y+\sqrt2\cos x\cos y\le\frac12$?

1. Put $a=\frac\pi2-x$, $b=\frac\pi2-y$. The condition becomes $\sin^2a+\sin^2b+\sqrt2\sin a\sin b\le\frac12$ on the same square $0\le a,b\le\frac\pi2$.
2. If $a+b=\frac\pi4$, take a triangle with angles $a,\,b,\,135^\circ$. By the Law of Sines its sides are proportional to $\sin a,\sin b,\sin135^\circ=\frac{\sqrt2}2$, and the Law of Cosines gives $\sin^2a+\sin^2b-2\sin a\sin b\cos135^\circ=\frac12$, i.e. equality holds on the line $a+b=\frac\pi4$.
3. The left side increases when $a$ or $b$ increases, so the region is exactly $a+b\le\frac\pi4$, a right triangle with legs $\frac\pi4$.
4. Area $=\dfrac12\left(\dfrac\pi4\right)^2=\boxed{\dfrac{\pi^2}{32}}$.