IMC 2006 · Day 2 · P9

Let $f(x) = \tan(\sin x) - \sin(\tan x)$. Then $$f'(x) = \frac{\cos x}{\cos^2(\sin x)} - \frac{\cos(\tan x)}{\cos^2 x} = \frac{\cos^3 x - \cos(\tan x) \cdot \cos^2(\sin x)} {\cos^2 x \cdot \cos^2(\tan x)}$$ Let $0 < x < \arctan \frac{\pi}{2}$. It follows from the concavity of cosine on $\left( 0, \frac{\pi}{2} \right)$ that $$\sqrt[3]{\cos(\tan x) \cdot \cos^2(\sin x)} < \frac{1}{3} \left[ \cos(\tan x) + 2 \cos(\sin x) \right] \le \cos \left[ \frac{\tan x + 2 \sin x}{3} \right] < \cos x\ ,$$ the last inequality follows from $$\left( \frac{\tan x + 2 \sin x}{3} \right)' = \frac{1}{3} \left[ \frac{1}{\cos^2 x} + 2 \cos x \right] \ge \sqrt[3]{\frac{1}{\cos^2 x} \cdot \cos x \cdot \cos x} = 1.$$ This proves that $\cos^3 x - \cos(\tan x) \cdot \cos^2(\sin x) > 0$, so $f'(x) > 0$, so $f$ increases on the interval $\left[ 0, \arctan \frac{\pi}{2} \right]$.

To end the proof it is enough to notice that (recall that $4 + \pi^2 < 16$) $$\tan \left[ \sin \left( \arctan \frac{\pi}{2} \right) \right] = \tan \frac{\pi/2}{\sqrt{1 + \pi^2/4}} > \tan \frac{\pi}{4} = 1\ .$$ This implies that if $x \in \left[ \arctan \frac{\pi}{2}, \frac{\pi}{2} \right]$ then $\tan(\sin x) > 1$ and therefore $f(x) > 0$.