$\tan x>x+\frac {x^3}3$ 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 ?
 A: Let $f(x)=\tan{x}-x-\frac{x^3}{3}.$
Now, $$f'(x)=\frac{1}{\cos^2x}-1-x^2=\tan^2x-x^2=(\tan{x}-x)(\tan{x}+x)>0.$$
Can you end it now?
A: At the point 
$$
\frac{1}{\cos^2x}\overset{\Large{?}}>1+x^2\iff\frac12(1+x^2)(1+\cos(2x))\overset{\Large{?}}<1
$$
one can insert the power series of the cosine to get
$$
\frac12(1+x^2)(1+\cos(2x))=(1+x^2)\left(1-x^2+\tfrac13x^4-\tfrac{2}{45}x^6\pm\cdots+\tfrac{(-1)^n}{(2n)!}(2x)^{2n}\right)
\\
=1-\tfrac23x^4+\tfrac{13}{45}x^6\mp\cdots+\tfrac{(-1)^n}{(2n+2)!}[(n+\tfrac12)(n+1)-1](2x)^{2n+2}
$$
which is an alternating series with continuously falling coefficients, so that the Leibniz test can be applied for $|x|<1$, giving that the first partial sum, consisting of the first term $1$, is an upper bound.
A: Due to the alternating nature of the sine and cosine series the partial sums are alternatingly upper and lower bounds (which is a corollary of the proof of the Leibniz test). Thus for $x\in (0,\frac\pi2)$, or more general $x\in(0,\sqrt6)$,
$$
x-\frac{x^3}6\le \sin x\le x\\
1-\frac{x^2}2\le \cos x\le 1-\frac{x^2}2+\frac{x^4}{24}
$$
Thus
\begin{align}
\sin x-(x+\frac{x^3}3)\cos x&\ge x-\frac{x^3}6-(x+\frac{x^3}3)(1-\frac{x^2}2+\frac{x^4}{24})
\\
&=\frac{x^5}8-\frac{x^7}{72}>\frac{x^5}{24}>0
\end{align}
proving the claimed inequality directly.
A: 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 $x\in\left(0,\frac{\pi}{2}\right)$.
