Prove $\sin(x)\tan(x) > x^2$ for $x \in ( \,0, \frac{\pi}{2}) \,$ Prove $\sin(x)\tan(x) > x^2$ for $x \in ( \,0, \frac{\pi}{2}) \,$
So I did the following:
Let $f(x) = \sin(x)\tan(x) - x^2$.
Then of course $f(0) = 0$ and I want to show that for $x \in ( \,0, \frac{\pi}{2}) \,$ $f'(x) > 0$ but I don't really know where to go from here as I can't get anything reasonable done with the derivative.
I was thinking of approximating it using the following inequality $\sin(x) < x < \tan(x)$. 
Is that a good direction?

If you can use Taylor series, then you may proceed as follows:


*

*$\sin(x)\tan(x) > x^2 \Leftrightarrow \boxed{\left(\frac{\sin x}{x}\right)^2 > \cos x}$ on $(0,\frac{\pi}{2})$
Using Taylor you get on $(0,\frac{\pi}{2})$:


*

*$\frac{\sin x}{x} > 1-\frac{x^2}{6} \Rightarrow \boxed{\left(\frac{\sin x}{x}\right)^2 > 1-\frac{x^2}{3} + \frac{x^4}{36}}$

*$1-\frac{x^2}{2} + \frac{x^4}{24} > \cos x$
Remains to show $1-\frac{x^2}{3} + \frac{x^4}{36} > 1-\frac{x^2}{2} + \frac{x^4}{24}$ on $(0,\frac{\pi}{2})$:
\begin{eqnarray*} 
& 1-\frac{x^2}{3} + \frac{x^4}{36} > 1-\frac{x^2}{2} + \frac{x^4}{24} &\\
& \stackrel{x > 0}{\Leftrightarrow} & \\
& |x| < 2\sqrt{3} &
\end{eqnarray*}
Since $\frac{\pi}{2} < 2\sqrt{3}$, we are done.
