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!

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.
