# How can I prove $2\sin x+\tan x \geq 3x$?

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.

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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, $$2\cos x+\sec^2 x-3 \geq 0 ~\forall x \in (0,\frac{\pi}{2})$$
Hint: Since $f^{\prime}(0)=0$, take $f^{\prime\prime}$ and so $f^{\prime\prime}(x)\ge 0$. This should be easier because you won't have the constant term
Indeed, $$f^{\prime}(x)=2\cos x+\frac{1}{\cos^2 x}-3$$ Thus, $$f^{\prime\prime}(x)=-2\sin x+\frac{2\sin x}{\cos^3 x}=2\sin x\frac{1-\cos^3 x}{\cos^3 x}> 0$$ since $\cos^3 x< \cos x< 1$ in $(0,\frac{\pi}2)$