<s>[INST] 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!
 [/INST] 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}
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