<s>[INST] For $0 < x < \pi/2 $, prove that $x + \frac{x^{3}}{3} <\tan x $ I proved as follows:

Let $f(x) = \tan x -  x - \frac{x^{3}}{3}.$
Then $f '(x) = \sec^{2}x - 1 - x^{2}$ and $f''(x) = 2\sec^{2}x\tan x-x > \tan x - x,$ since $\\tan x > x,$ $f''(x) > 0$ so $f'(x)$ is increasing.
Since $f '(0) = 0,$ $f '(x) > 0$ for all $x \in (0, \frac{\pi}{2}).$
So $f(x)$ is increasing. since $f(0) = 0,$ $f(x) > 0$ for all $x \in (0, \frac{\pi}{2}).$

But my proof seems complex. Is there any simpler way to prove this?
 [/INST] Differentiate the given inequation three times 
$$1+x^2\stackrel?<\tan^2x+1,$$
$$x\stackrel?<\tan x(\tan^2x+1),$$
$$1\stackrel?<(\tan^2x+1)^2+2\tan^2x(\tan^2x+1).$$
The final one is obviously true and equality holds at $x=0$, so that the second holds, by integration. That one holds at $x=0$, so that the first holds. And as the first also holds at $x=0$, so does the given one.

You can generalize the reasoning with more terms of the Taylor expansion.
</s>