Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 10 down vote accepted

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, $$2\cos x+\sec^2 x-3 \geq 0 ~\forall x \in (0,\frac{\pi}{2})$$

share|cite|improve this answer
This answer is very good. Thank you. – Argha Dec 24 '12 at 6:01

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 $(0,\frac{\pi}2)$

share|cite|improve this answer
You answer is also helpful.Thank you. – Argha Dec 24 '12 at 6:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.