<s>[INST] 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 $1$?
 [/INST] An alternative proof for the desired inequality without using
Taylor series would be to notice that 
$$
 f(x) = \frac 12 (\sin x + \tan x)
$$
satisfies $f(0) = 0$ and 
$$
 f'(x) = \frac 12  \left(\cos x + \frac{1}{\cos^2 x} \right) \ge
 \sqrt{\cos x \cdot \frac{1}{\cos^2 x}} = \frac{1}{\sqrt{\cos x}}
 \ge 1
$$
for $0 \le x < \frac \pi 2$.
</s>