$\tan x>x+\frac {x^3}3$ for $x\in(0,\frac\pi2)$ 
$$\tan x>x+\frac {x^3}3~\text{ for }~x\in\left(0,\frac\pi2\right)$$

My solution:
Both functions are monotone, increasing, equal at $x=0$. If i could show that the derivative of the first is greater than then derivative of second function that would be it. Taking the derivatives:
$$\frac1{\cos^2 x}>1+x^2$$
Applying the same reasoning on the derivatives  and taking their derivatives we get
$$\frac{\tan x}{\cos^2 x}>x$$
Doing the same thing (it is possible to stop here if we use $\tan x>x$)
$$\frac{1+\tan x\sin 2x}{\cos^4 x}>1$$
It is easy to see that last inequality is true therefore all previous are also true.
Is my solution correct, i would be very disappointed if it weren't. Are there different ways to solve this ?

Since $\tan(x)=-\frac{d}{dx}\log\cos x$ and 
$$ \cos(x)=\prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2 \pi^2}\right) $$
we have
$$ \frac{\tan x}{x}=\sum_{n\geq 0}\frac{8}{(2n+1)^2 \pi^2-4x^2}=\sum_{n\geq 0}\frac{8}{(2n+1)^2 \pi^2}\sum_{m\geq 0}\left(\frac{4x^2}{\pi^2(2n+1)^2}\right)^m $$
and by exchanging $\sum_{n\geq 0}$ and $\sum_{m\geq 0}$ we get
$$ \color{red}{\frac{\tan x}{x}} = \sum_{m\geq 0}\frac{4^{m+1} x^{2m}}{\pi^{2m+2}}\cdot 2\sum_{n\geq 0}\frac{1}{(2n+1)^{2m+2}}=\color{red}{\sum_{m\geq 0}2(4^{m+1}-1)x^{2m}\cdot \frac{\zeta(2m+2)}{\pi^{2m+2}}} $$
and the given inequality follows from $\zeta(2)=\frac{\pi^2}{6}$, $\zeta(4)=\frac{\pi^4}{90}$ and $\zeta(2m+2)\geq 1$ for any $m\in\mathbb{N}$.
This actually leads to
$$ \frac{\tan x}{x} \geq 1 + \frac{x^2}{3} + \frac{6 x^4 \left(21 \pi ^2-20 x^2\right)}{4 \pi ^4 x^4-5 \pi ^6 x^2+\pi ^8}$$
for any $x\in\left(0,\frac{\pi}{2}\right)$.
