<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] If we can show that $f$ is decreasing on $(0,\frac{\pi}{2}]$, then it is easy to finish by checking the endpoints.
To show that $f$ is decreasing on this interval, it is enough to show that $f'$ is negative on this interval.  You computed
$$f'(x)=\frac{ x^2\sin x-(1-\cos x)x-\sin x(1-\cos x)}{(x-\sin x)^2}$$
which is negative if and only if the numerator is negative.
For convenience, we will rewrite the numerator as
$$x^2\sin x-(1-\cos x)(x+\sin x).$$
We can bound the numerator by using that fact that Taylor series for sine and cosine are alternating series, and therefore truncating it gives us bounds.
For $x\in(0,\frac{\pi}{2}]$, we have
$$x^2\sin x\le x^2(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!})$$
$$(1-\cos x)(x+\sin x)\ge(1-(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}))(x+(x-\frac{x^3}{3!}))$$
where for the second inequality we use that both factors on the right hand side are positive on this interval.  Therefore,
\begin{align*}
&x^2\sin x-(1-\cos x)(x+\sin x)\\
\le &x^2(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}) - (1-(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}))(x+(x-\frac{x^3}{3!}))\\
=&-\frac{x^7((x^2-30)^2+108)}{725760}
\end{align*}
and we can see that the right hand side is always negative on the interval.
(How did I know how far to expand each Taylor series?  Trial and error.  There might be some way to see how far to expand each using the fact that they Taylor series for the numerator of $f'(x)$ is $O(x^7)$, but I don't see how at the moment.)
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