How to prove the lower bound of $\frac{x^2}{\sin^2x}$? How to prove $$1+\frac{x^2}{3}\leq \frac{x^2}{\sin^2x}, x\in (0,\pi/2)?$$
I do want to show it by intermediate value theorem as
$$\frac{1}{\sin^2x}-\frac{1}{x^2}
=\frac{2}{\xi^3}(x-\sin x)>\frac{2(x-\sin x)}{x^3}.$$
However, this may corrupt, since the rhs $<=\frac{1}{3}$.

You might prefer to prove
$$ 1+\frac{x^2}{3} \le e^{x^2/3} \le \left(\frac{x}{\sin x}\right)^2 $$
For the second inequality, the slickest proof is perhaps
$$ \frac{\sin x}{x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{\pi^2 n^2}\right)
\le \prod_{n=1}^\infty \exp\left(-\frac{x^2}{\pi^2 n^2}\right)
= \exp\left(-\frac{x^2}{\pi^2} \sum_{n=1}^\infty \frac1{n^2}\right)
= e^{-x^2/6}
$$
but this requires a couple big guns in the first and last steps.  A more elementary, but tedious, proof can be had by comparing power series:
$$ \frac{\sin x}{x} \le 1 - \frac{x^2}{6} + \frac{x^4}{120}
\le 1 - \frac{x^2}{6} + \frac{x^4}{72} - \frac{x^6}{1296}
\le e^{-x^2/6} $$
The first and last inequalities here come from the usual bounds for alternating series when the omitted terms decrease in absolute value (which they do here if $x$ is not very big); the middle inequality is routine algebra, and is valid for $|x|\le\frac6{\sqrt5}$, which covers the interval you want.

