Prove $\frac{2 \sin x}{3}+\frac{\tan x}{3} > x$ for $x \in (0, \frac{\pi}{2})$ Please, help
Prove $\frac{2 \sin x}{3}+\frac{\tan x}{3}  > x$
$x \in (0, \frac{\pi}{2})$

Letting $f(x)=2\sin x+\tan x-3x$, then differentiate it. You'll get the answer.
$$f^{\prime}(x)=2\cos x+\frac{1}{\cos^2 x}-3=\frac{2\cos^3 x-3\cos^2 x+1}{\cos^2 x}=\frac{(\cos x-1)^2(2\cos x+1)}{\cos^2 x}$$
