prove $(\sin x)^{-2}-x^{-2}\leq 1-\frac{4}{{\pi}^{2}},x\in(0,\pi/2]$ $(\sin x)^{-2}-x^{-2}\leq 1-\frac{4}{{\pi}^{2}},x\in(0,\pi/2]$
How to deal with this problem?
Observing that when $x=\pi/2$, the above inequality becomes equality.
Firstly, denote $f(x)=(\sin x)^{-2}-x^{-2}$ and then take derivative of $f(x)$. We have
$$-2(\sin x)^{-3}\cos x+2x^{-3}$$ Next, how to analysis the sign of $f'(x)$?
Any hints are wellcome! Thanks!
 A: It's related to :
$$|\frac{1}{\sin(x)}-\frac{1}{x}|\leq 1-\frac{2}{\pi}$$
Called Prestin's inequality
Use the fact :
$$\left|\frac{1}{\sin^{2}(x)}-\frac{1}{x\sin\left(x\right)}\right|-\left(1-\frac{2}{\pi}\right)\geq (\sin x)^{-2}-x^{-2}-\left(1-\frac{4}{\pi^{2}}\right)$$
And :
$$\left|\frac{1}{\sin(x)}-\frac{1}{x}\right|-\left|\sin\left(x\right)\right|\left(1-\frac{2}{\pi}\right)\leq 0$$
Or using Jordan's inequality :
$$\left|\frac{1}{\sin(x)}-\frac{1}{x}\right|-\left|\frac{2}{\pi}x\right|\left(1-\frac{2}{\pi}\right)\leq 0$$
Or using Taylor's series for $x\in(0,1]$ :
$$\left|\frac{1}{x-\frac{1}{6}x^{3}}-\frac{1}{x}\right|-\left|\frac{2}{\pi}x\right|\left(1-\frac{2}{\pi}\right)\leq 0$$
Or :
$$\frac{\left|x\right|\left(\frac{\pi^{2}}{\left|x^{2}-6\right|}-2\pi+4\right)}{\pi^2}\leq 0$$
Now on $x\in[1,\pi/2]$ :
$$\left|\frac{1}{1-1/2(x-\pi/2)^{2}}-\frac{1}{x}\right|-\left|\frac{2}{\pi}x\right|\left(1-\frac{2}{\pi}\right)\leq 0$$
Or :
$$-\frac{\left((2x-\pi)(4\pi x^{3}-8x^{3}-2\pi^{2}x^{2}+4\pi x^{2}+2\pi^{2}x-8\pi x+16x-\pi^{3}+8\pi)\right)}{(\pi^{2}x(4x^{2}-4\pi x+\pi^{2}-8))}\leq 0$$
A: Let $f(x)=\sin^{-2}x-x^{-2}$.
Note
$$ f'(x)=\frac{2}{x^3}-\frac{2\cos x}{\sin^3x}=\frac{2(\sin^3x-x^3\cos x)}{x^3\sin^3x}=\frac{g(x)}{x^3\sin^3x}. $$
where
$$ g(x)=\sin^3x-x^3\cos x.$$
Now
$$ g'(x)=x^3\sin x-3\cos x(x^2-\sin^2x)\ge0, \forall x\in(0,\frac\pi2] $$
which implies that $g(x)$ is increasing and hence
$$ g(x)> g(0)=0, \forall x\in(0,\frac\pi2]. $$
This also implies that $f(x)$ is increasing. So $x\in[0,\frac\pi2]$
$$ f(x)\le f(\frac\pi2) $$
or
$$(\sin x)^{-2}-x^{-2}\leq 1-\frac{4}{{\pi}^{2}},x\in(0,\pi/2]. $$
$\bf{Update:}$ Actually we don't need $g'(x)\ge0$. To prove $g(x)\ge 0$ for $x\in(0,\pi/2]$, using
$$ \sin x\ge x-\frac{x^3}6, \cos x\le 1-\frac{x^2}2+\frac{x^4}{24}, $$
one has
\begin{eqnarray}
g(x)&=&x^3\bigg[\bigg(\frac{\sin x}{x}\bigg)^3-\cos x\bigg]\\
&=&x^3\bigg[\bigg(1-\frac{x^2}6\bigg)^3-\bigg(1-\frac{x^2}2+\frac{x^4}{24}\bigg)\bigg]\\
&=&\frac1{216}x^7(9-x^2)>0, x\in(0,\pi/2]. 
\end{eqnarray}
A: The Laurent series expansion of $\frac{1}{\sin x} $, valid for $0<|x|< \pi$, is
$$\frac{1}{\sin x} = \frac{1}{x} + \frac{1}{6} x + \frac{7}{360} x^3 + \frac{31}{1520} x^5 + \cdots $$
with all coefficients positive. We conclude that also $\frac{1}{\sin^2 x}$ has a Laurent expansion with positive coefficients valid for $0< |x| < \pi$
$$\frac{1}{\sin^2 x} = \frac{1}{x^2} + \frac{1}{3}+ \frac{1}{15} x^2 + \frac{2}{189} x^4 + \frac{1}{675} x^6 + \cdots$$
From this several inequalities follow.
