TITLE: How can I prove $2\sin x+\tan x \geq 3x$?
QUESTION [11 upvotes]: Prove that $$2\sin x+\tan x \geq 3x,\quad 0 < x< \frac{\pi}{2}$$

Trial: $2\sin x+\tan x \geq 3x\equiv 2\sin x+\tan x -3x\geq 0$. So, let $f(x)=2\sin x+\tan x-3x$.Here $f(0)=0$ and If I can show $f'(x) \geq 0,\forall x \in (0,\frac{\pi}{2})$, then I can prove the inequality. Now $f'(x)=2\cos x + \sec^2x-3$.How to show $f'(x) \geq 0$. Please help.

REPLY [15 votes]: Apply $\text{AM} \geq \text{GM}$ with $\cos x, \cos x, \sec^2 x$ We get $$(\frac{\cos x+\cos x+\sec^2 x}{3}) \geq (\cos x.\cos x.\sec^2 x)^{1/3}=1$$ So, $$2\cos x+\sec^2 x-3 \geq 0 ~\forall x \in (0,\frac{\pi}{2})$$