Prove inequality between functions

Question

Show that $x^2<\tan{x}\arctan{x}$ if $x\in(0;\pi/2)$. I have thought about proving that $f(x) = \tan{x}\arctan{x} - x^2$ is monotonically increasing(its derivative is larger than 0), but its derivative ($f'(x) = \frac{\arctan{x}}{\cos^2x} + \frac{\tan x}{1 + x^2}-2x$) is hard to compare with 0. Please, can you give me a hint?

Answer 1

Since $\tan x > x$ for all $x \in \left( 0,\dfrac{\pi}{2} \right)$, $\dfrac{\tan(x)}{x}> 1$. We claim that $g(x) = \dfrac{\tan(x)}{x}$ is strictly increasing on that interval.

$$g'(x) = \left(\frac{\tan x}x\right)'=\frac{\frac{x}{\cos^2 x}-\tan x}{x^2}=\frac{x-\sin x\cos x}{x^2\cos^2 x}=\frac{2x-\sin(2x)}{2x^2\cos^2x}>0,$$ Source: linked answer

Consider $g(\arctan x) < g(x)$, and you'll get the desired inequality.

$$\frac{x}{\arctan x}<\frac{\tan x}{x} \iff \tan x \arctan x > x^2$$

Answer 2

Put it this way: you have to prove that

$$\frac{x}{\arctan x}<\frac{\tan x}{x}$$

Introduce:

$$x=\tan y$$

...and notice that $x>y$.

With this in mind you have to prove that:

$$\frac{\tan y}{y}<\frac{\tan x}{x}$$

In other words, you have to prove that $\frac{\tan x}{x}$ is strictly increasing. The first derivative is:

$$(\frac{\tan x}{x})'=\frac{2x-\sin 2x}{2x^2}$$

You should be able to draw the right conclusion from here.