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?
 A: Clearly we can see that
$$
\tan(x) ,\cos(x)>0 \, \forall x \in (0,\frac{ \pi}{2}) \, .
$$
Now as you said, the derivative is
$$
f'(x)=\cos(x)\tan(x)+ \frac{\sin(x)}{\cos^2(x)} -2x \\=\cos(x)\tan(x)+\tan(x)\sec(x)-2x \\
=\tan(x)\left(\cos(x)+\frac{1}{\cos(x)}\right)-2x \geq 0
$$
since 
$$
\tan(x)>x ,x+\frac{1}{x} \geq 2 \,  \forall x>0 \, .
$$
A: Using Geometric  mean $\geq $ Harmonic mean
$$\sqrt{\sin x\cdot \tan x}\geq \frac{2}{\frac{1}{\sin x}+\frac{1}{\tan x}}=2\tan \frac{x}{2}>x$$
becsuse $\displaystyle \tan x>x\;\forall \;x\in(0,90^\circ)$
So $$\sin x\cdot \tan x>x^2\;\forall\; x\in (0,90^\circ)$$
A: Let $f(x)=\sin(x)\tan(x)-x^2=\frac{1}{\cos(x)}-\cos(x)-x^2$. Then $f(0)=0$, $f'(0)=0$ and for $x\in (0,\pi/2)$,
$$f''(x)=\underbrace{\frac{2\sin^2(x)}{\cos^3(x)}}_{>0}+\underbrace{\frac{1}{\cos(x)}+\cos(x)}_{> 2}-2>0.$$
Then the desired inequality follows easily: $f$ is strictly convex over $(0,\pi/2)$ and therefore its graph lies above the tangent line at $x=0$, i.e. the line $y=0$.
A: 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.
A: A possible approach: Use Taylor's theorem to show that
$$
 \sin x > x - \frac{x^3}{6} \\
 \tan x > x + \frac{x^3}{3}
$$
for $0 < x < \frac\pi 2$. Then 
$$
\sin x \tan x > x^2 + \frac{x^4}{6} - \frac{x^6}{18}
 = x^2 + \frac{x^4(3 - x^2)}{18} > x^2
$$
in that interval.
