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