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?

For $0 < x < \pi/2$ we have $\tan'(x) = 1 + \tan^2(x)$ and therefore
$$
 \tan(x) = \int_0^x (1 + \tan^2(u)) \, du > \int_0^x  1 \, du = x \, .
$$
It follows that
$$
\tan(x) = \int_0^x (1 + \tan^2(u)) \, du > \int_0^x (1 +u^2) \, du = x + \frac 13 x^3 \, .
$$
This process can be repeated to get better lower bounds:
$$
\tan(x) = \int_0^x (1 + \tan^2(u)) \, du >  \int_0^x (1 + (u + \frac {u^3}3)^2) \, du = x + \frac 13 x^3 + \frac{2}{15} x^5 + \frac{1}{54} x^7 \, .
$$
