Prove that all odd derivate of $\tan(x)$ at $x=0$ is at least $1$. This was the original excercise:

Prove that $$\frac{\sin{x}+\tan{x}}{2} \geq x$$ where $x \in (0,\frac{\pi}{2})$

This is how I did it:
Knowing that 
\begin{align}\sin{0}&=0\\
\cos{0}&=1\\
\frac{\mathrm{d} \sin{x}}{\mathrm{d} x}&=\cos{x}\\
\frac{\mathrm{d} \cos{x}}{\mathrm{d} x}&=- \sin{x}\end{align}
and the Taylor series of $f(x)$ is: $$\sum_{n=0}^{\infty} \frac{x^n\frac{\mathrm{d}^n f(0)}{\mathrm{d} x^n}}{n!}$$
we can easily see that the Taylor series of $\sin(x)$ is:
$$\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$$
We know what the tan(x) function is odd, so the n-th derivate will be $0$ for all even $n$. For odd numbers, it will be $1,2,16,272,\cdots$ So if we sum the $2$ Taylor series, we will get this:
$$\sin{x}+\tan{x}=2x+\frac{x^3}{3!} (-1+1)+\frac{x^5}{5!}(16+1)+\frac{x^7}{7!}(272-1)+...$$
Where we can see that all part will be positive (or $0$) on the given interval. Back to the original inequality:
$$2x+\frac{x^3}{3!} (-1+1)+\frac{x^5}{5!}(16+1)+\frac{x^7}{7!}(272-1)+...\geq 2x$$
$$\frac{x^3}{3!} (-1+1)+\frac{x^5}{5!}(16+1)+\frac{x^7}{7!}(272-1)+... \geq 0$$
Which is true in the  given interval.
But how could I prove that all odd derivate of $\tan(x)$ at $x=0$ is  at least $1$?

Nice exercise. We have that $f(x)=\tan(x)$ fulfills the differential equation
$$ f'(x) = 1 + f(x)^2 \tag{1}$$
with $f(0)=0$ and $f'(0)=1$. The even derivatives at the origin are obviously zero, since $\tan(x)$ is an odd function, and the odd derivatives can be computed by applying $\frac{d^{2k}}{dx^{2k}}$ to both sides of $(1)$.
For instance
$$ f'''(x) = 2\,f'(x)^2+2\,f(x)\,f''(x)\tag{2}$$
so, by induction, the odd derivatives of $\tan(x)$ at the origin are positive integers.

An alternative derivation of tangent numbers comes from
$$ \cos(x)=\prod_{m\geq 0}\left(1-\frac{4x^2}{(2m+1)^2\pi^2}\right). \tag{3}$$
By applying $-\frac{d}{dx}\log(\cdot)$ to both sides of $(3)$,
$$ \frac{\tan(x)}{x}=\sum_{m\geq 0}\frac{2}{\left(m+\frac{1}{2}\right)^2 \pi^2-x^2}=\sum_{k\geq 0}\frac{2x^{2k}}{\pi^{2k+2}}\sum_{m\geq 0}\frac{1}{\left(m+\frac{1}{2}\right)^{2k+2}}\tag{4}$$
hence:
$$ \frac{\tan(x)}{x}=\sum_{k\geq 0}\frac{2 x^{2k}(2^{2k+2}-1)\zeta(2k+2)}{\pi^{2k+2}}\tag{5}$$
and it is enough to show that the denominator of $\frac{\zeta(2k+2)}{\pi^{2k+2}}\in\mathbb{Q}$ is a divisor of $(2k+1)!$. That directly follows from the Fourier series of Bernoulli polynomials.

About the origin exercise, you may simply notice that
$$ \frac{\sin(x)+\tan(x)}{2} = \frac{1}{2}\int_{0}^{x}\left(\cos(t)+\frac{1}{\cos^2(t)}\right)\,dt> \frac{1}{2}\int_{0}^{x}2\,dt = x $$
since $u+\frac{1}{u^2}> 2$ for any $u\in(0,1)$. The similar inequality
$$ \forall x\in\left(0,\frac{\pi}{2}\right),\qquad 2\sin(x)+\tan(x) > 3x $$
was proved in 1631 by Snellius.
