The given inequality is successively equivalent to
$$8 \cos x \cos y (\cos x \cos y + \sin x \sin y) + 1 > 4 \cos^2 x + 4 \cos^2 y,$$
$$8 \cos^2 x \cos^2 y - 4 \cos^2 x - 4 \cos^2 y + 2 \sin 2x \sin 2y + 1 > 0,$$
$$2 (2 \cos^2 x - 1) (2 \cos^2 y - 1) + 2 \sin 2x \sin 2y - 1 > 0,$$
$$\cos 2x \cos 2y + \sin 2x \sin 2y > \frac{1}{2},$$
$$\cos (2x - 2y) > \frac{1}{2}.$$
By the pigeonhole principle, in one of the sets
$$\langle 0, \frac{\pi}{6} \rangle, \quad \left[ \frac{\pi}{6}, \frac{\pi}{3} \right), \quad \left[ \frac{\pi}{3}, \frac{\pi}{2} \right]$$
there are two out of four given numbers. Let these be $x$ and $y$.
Now we have $|2x - 2y| < \frac{\pi}{3}$ and $\cos(2x - 2y) > \frac{1}{2}$, which proves the claim.