Iberoamerican Mathematical Olympiad

Solution:

We have $\sin 2x + \sin 2y + \sin 2z - 2 \sin x \cos y - 2 \sin y \cos z = 2 \sin x (\cos x - \cos y) + 2 \sin y (\cos y - \cos z) + 2 \sin z \cos z$, so we wish to show that

$\sin x (\cos x - \cos y) + \sin y (\cos y - \cos z) + \sin z \cos z < \pi / 2$ (*).

We have to consider six cases:

(1) $x \leq y \leq z$;

(2) $x \leq z \leq y$;

(3) $y \leq x \leq z$;

(4) $y \leq z \leq x$;

(5) $z \leq x \leq y$;

(6) $z \leq y \leq x$.

The first case is obvious from the diagram, because the lhs represents the shaded area, and the rhs represents the whole quarter circle.

In cases (2) and (5) the second term is negative, and $-\sin y < -\sin x$, so the sum of the first two terms is less than $\sin x (\cos x - \cos y) + \sin x (\cos y - \cos z) = \sin x (\cos x - \cos z)$. But by the same argument as the first case the two rectangles represented by $\sin x (\cos x - \cos z)$ and $\sin z \cos z$ are disjoint and fit inside the quarter circle. So we have proved (2) and (5).

In cases (3) and (4), the first term is negative. The remaining two terms represent disjoint rectangles lying inside the quarter circle, so again the inequality holds.

In case (6) the first two terms are negative. The last term is $1/2 \sin 2z \leq 1/2 < \pi / 2$, so the inequality certainly holds.