Some algebraic manipulation yields: $8\cos(x)\cos(y)\cos(x-y) + 1 > 4(\cos^{2}(x)+\cos^{2}(y))$ $8\cos(x)\cos(y)[\cos(x)\cos(y) + \sin(x)\sin(y)] + 1 > 4\cos^{2}(x)+4\cos^{2}(y)$ $8\cos^2(x)\cos^2(y) - 4\cos^{2}(x) - 4\cos^{2}(y) + 2 + 8\cos(x)\cos(y)\sin(x)\sin(y) > 1$ $2(2\cos^2(x)-1)(2\cos^2(y)-1) + 2(2\sin(x)\cos(x))(2\sin(y)\cos(y)) > 1$ $2\cos(2x)\cos(2y) + 2\sin(2x)\sin(2y) > 1$ $\cos(2x - 2y) > \dfrac{1}{2} = \cos\left(\dfrac{\pi}{3}\right)$ Since $x,y \in \left(0,\dfrac{\pi}{2}\right)$, this is equivalent to: $|2x-2y| < \dfrac{\pi}{3} \Longleftrightarrow |x-y| < \dfrac{\pi}{6}$ Let the four distinct numbers be $0 < a < b < c < d < \dfrac{\pi}{2}$. Suppose that no two of $a,b,c,d$ satisfy $|x-y| < \dfrac{\pi}{6}$. Then we have: $|b-a| \ge \dfrac{\pi}{6} \Rightarrow b \ge a + \dfrac{\pi}{6}$ $|c-b| \ge \dfrac{\pi}{6} \Rightarrow c \ge b + \dfrac{\pi}{6} \ge a + \dfrac{\pi}{3}$ $|d-c| \ge \dfrac{\pi}{6} \Rightarrow d \ge c + \dfrac{\pi}{6} \ge a + \dfrac{\pi}{2}$ However, if $a > 0$, then $d > \dfrac{\pi}{2}$, a contradiction. Therefore, two of $a,b,c,d$ satisfy $|x-y| < \dfrac{\pi}{6}$, and thus, the original inequality.