How prove this inequality $\tan{(\sin{x})}>\sin{(\tan{x})}$

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 $\lim_{x\to 0^{+}}\dfrac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}$

Solution 1:

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, $$\frac{1}{\theta}\int_{0}^{\theta}\left(\frac{1}{\cos^2 u}+2\cos u\right)\,du\geq 3.$$