16
$\begingroup$

How prove this inequality $$\tan{(\sin{x})}>\sin{(\tan{x})},0<x<\dfrac{\pi}{2}$$

this PDF give a ugly methods : http://wenku.baidu.com/link?url=CHnWPdmjsqSmNAQhL4bOmDfUVc0Tc5nWCBQWNB1lweG-LBnIlQWje_qdAUBQgUQh3C6znVCpIoefzzNvgfTMv8xrw8jd2sZy_Mlgy-dJKo3

I post this solution:let

$$f(x)=\tan{(\sin{x})}-\sin{(\tan{x})}$$ then we have $$f'(x)=sec^2{(\sin{x})}\cos{x}-\cos{(\tan{x})}\sec^2{x}=\dfrac{\cos^3{x}- \cos{(\tan{x})}\cos^2{(\sin{x})}}{\cos^2{(\sin{x})}\cos^2{x}}$$ case 1:$0<x<\arctan{\dfrac{\pi}{2}}$, then we have $$0<\tan{x}<\dfrac{\pi}{2},0<\sin{x}<\dfrac{\pi}{2}$$ so Use AM-GM inequality we have $$\sqrt[3]{\cos{(\tan{x})}\cos^2{(\sin{x})}}\le\dfrac{1}{3}[\cos{(\tan{x})}+2\cos{(\sin{x})}]\le\cos{\dfrac{\tan{x}+2\sin{x}}{3}}$$ use $$\tan{x}+2\sin{x}>3x$$ so $$f'(x)>0\Longrightarrow f(x)>0$$ case2: $\arctan{\dfrac{\pi}{2}}\le x\le\dfrac{\pi}{2}$, so $$\sin{(\arctan{\dfrac{\pi}{2}})}<\sin{x}<1$$ since $$\sin{(\arctan{\dfrac{\pi}{2}})}=\dfrac{\pi}{\sqrt{4+\pi^2}}>\dfrac{\pi}{4}$$ $$\Longrightarrow \dfrac{\pi}{4}<\sin{x}<1$$,so $$1<\tan{(\sin{x})}<\tan{1},$$,so $$f(x)>0$$

I think this inequality have other simple methods.Thank you

It is said can use integral inequality,But I can't

In fact,we have $$\tan{(\sin{x})}-\sin{(\tan{x})}=\dfrac{1}{30}x^7+o(x^7),x\to 0$$ can see:How find this limit $\lim_{x\to 0^{+}}\dfrac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}$

$\endgroup$
  • $\begingroup$ Will you award the bounty this time? $\endgroup$ – Jack D'Aurizio Mar 10 '14 at 10:37
  • $\begingroup$ Hello,It must!Thank you $\endgroup$ – math110 Mar 10 '14 at 11:49
10
+50
$\begingroup$

I think I got a very nice proof. Consider that: $$\tan(\sin x)=\int_{0}^{\sin x}\frac{d\theta}{\cos^2\theta}=\int_{0}^{\sin x}\frac{d\theta}{1-\sin^2\theta}=\int_{0}^{x}\frac{\cos\psi\,d\psi}{\cos^2(\sin\psi)},$$ while: $$\sin(\tan x)=\int_{0}^{\tan x}\cos\theta\,d\theta = \int_{0}^{x}\frac{\cos(\tan\psi)\,d\psi}{\cos^2\psi}.$$ Since for any $x>1$ we have: $$\tan(\sin x)>\tan(\sin 1)>\frac{19}{17},\qquad \sin(\tan x)\leq 1,$$ we just have to prove that for any $\theta\in[0,1]$ $$\cos^3\theta \geq \cos(\tan\theta)\cos^2(\sin\theta)\tag{1}$$ holds, or, for any $u\in[0,\tan 1]$: $$\frac{1}{(1+u^2)^{3/2}}\geq \cos(u)\cos^2\left(\frac{u}{\sqrt{1+u^2}}\right).\tag{2}$$ Since $\log\cos x$ is a concave function on that interval, we only need to prove: $$\forall u\in[0,\tan 1],\qquad (1+u^2)\cdot\cos^2\left(\frac{1}{3}\left(u+\frac{2u}{\sqrt{1+u^2}}\right)\right)\leq 1,$$ or: $$\forall \theta\in[0,1],\qquad \cos^2\left(\frac{\tan\theta+2\sin\theta}{3}\right)\leq \cos^2\theta,$$ $$ \forall \theta\in[0,1],\qquad \tan\theta + 2\sin\theta \geq 3\theta.\tag{3}$$ The last inequality is true for sure since, by the AM-GM inequality, $$\frac{1}{\theta}\int_{0}^{\theta}\left(\frac{1}{\cos^2 u}+2\cos u\right)\,du\geq 3.$$

$\endgroup$
  • 1
    $\begingroup$ Oh,It's nice ! Thank you $\endgroup$ – math110 Mar 10 '14 at 11:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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