<s>[INST] Prove using calculus or otherwise the following inequality Prove that$$ \frac {2}{\pi -2}\ \le\ \frac {\sin x-x\cos x}{x-\sin x}\ <\ 2$$where $$x\in\left(0,\frac{\pi}{2}\right]$$
My Attempt:$$ f'(x)=\frac{ x\sin x(x-\sin x)-(1-\cos x)(\sin x-x\cos x)}{(x-\sin x)^2}=\frac{ x^2\sin x-(1-\cos x)x-\sin x(1-\cos x)}{(x-\sin x)^2}$$
After this not able to get breakthrough.
 [/INST] I did by using brute force
$$f'(x)=\frac{\sin x\left(x-\frac{1-\cos x}{2\sin x}(1+\sqrt{5+4\cos x})\right)\left(x-\frac{1-\cos x}{2\sin x}(1-\sqrt{5+4\cos x})\right)}{(x-\sin x)^2}=\frac{\sin x\left(x-\tan \frac{x}{2}(1+\sqrt{5+4\cos x})\right)\left(x-\tan \frac{x}{2}(1-\sqrt{5+4\cos x})\right)}{(x-\sin x)^2}$$
Let $$g(x)=x-\tan \frac{x}{2}\left(1+\sqrt{5+4\cos x}\right)$$
$$g'(x)=1-\frac{1}{2}\sec^2\frac{x}{2}\left(1+\sqrt{5+4\cos x}\right)-\tan \frac{x}{2}.\frac{-4\sin x}{2\sqrt{5+4\cos x}}$$
$$=1-\frac{1+\sqrt{5+4\cos x}}{1+\cos x}+\frac{2(1-\cos x)}{\sqrt{5+4\cos x}}$$
$$=\frac{\cos x\left(\sqrt{5+4\cos x}-4\right)-2\cos^2x-3}{(1+\cos x)\sqrt{5+4\cos x}}<0$$
$g(x)$ is a decreasing function. So,if $x>0$,then $g(x)<g(0)$. i.e  $g(x)<0$
Similarly, Let $$h(x)=x-\tan \frac{x}{2}\left(1-\sqrt{5+4\cos x}\right)$$
$$h'(x)=1-\frac{1}{2}\sec^2\frac{x}{2}\left(1-\sqrt{5+4\cos x}\right)-\tan \frac{x}{2}.(-1)\frac{-4\sin x}{2\sqrt{5+4\cos x}}$$
$$=1-\frac{1-\sqrt{5+4\cos x}}{1+\cos x}-\frac{2(1-\cos x)}{\sqrt{5+4\cos x}}$$
$$=\frac{\cos x\left(\sqrt{5+4\cos x}+4\right)+2\cos^2x+8}{(1+\cos x)\sqrt{5+4\cos x}}>0$$
$h(x)$ is a increasing function. So,if $x>0$,then $h(x)>h(0)$. i.e  $h(x)>0$
Thus it can be concluded that $$f'(x)<0$$ and we are done.
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