Take the 2-minute tour ×
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.

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|improve this question

2 Answers 2

up vote 7 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|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|improve this answer
    
You answer is also helpful.Thank you. –  Argha Dec 24 '12 at 6:02

Your Answer

 
discard

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.