TITLE: Prove: $\sin (\tan x) \geq {x}$
QUESTION [26 upvotes]: I bumped into this question: 

Question: Prove that for $x\in \Bigl[0,\dfrac {\pi}{4}\Bigr]$, $$\sin (\tan x) \geq {x}$$

This seems to be an innocent inequality but I am already exhausted trying to prove it. I followed different lines of thought and mention $3$ of them below:
$1.$ I tried using $\sin x<x<\tan x$. Knowing that $\sin x$,$\tan x$ increase in the mentioned interval, I tried applying $\tan$ and $\sin$ on different sides of $\sin x<x<\tan x$ in order to achieve something like that in the question.. But I just could not come to the inequality in the question.
$2.$ I tried using the function $f(x)=\sin(\tan x)-x$. $f(0)=0$. If we prove that the derivative $f'(x)=(\sec ^2 x)(\cos(\tan x))-1$ is positive, then we will be done. But proving even this seems very difficult.
$3.$ Since $\arcsin x$ is one-one and increasing in the mentioned interval, we can take $\arcsin$ on both sides of the inequality $\sin (\tan x) \geq {x}$ to get $\tan x > \arcsin x$. Thus proving  $\tan x > \arcsin x$ would be equivalent to solving the problem.
All the three approaches are seeming impossible.. Any suggestions or other methods?
Thanks..

REPLY [15 votes]: From $(2)$ it boils down to show that $\cos(\tan(x)) \geq \cos^{2}(x)$. And note that $\cos(x) > 1-\dfrac{1}{2}x^{2}$ which gives that $\cos(\tan(x)) \geq 1-\frac{1}{2}\tan^{2}(x)$. Now it suffices for us to show $$1-\frac{1}{2}\tan^{2}(x) 
\geq \cos^{2}(x) \qquad \forall \ x \in [0,\frac{\pi}{4}]$$ This can be re-written as $$2\cos^{4}(x) - 3 \cos^{2}(x) + 1 \leq 0$$ or $$(2\cos^2x -1)(\cos^2x-1)\leq0$$ which actually holds for all $x \in \Bigl[0,\dfrac{\pi}{4}\Bigr]$.

REPLY [4 votes]: Another method:
Using $\sin(x)=x-\dfrac{x^3}{3!}+\ldots$  we see that it suffices to prove:
$\tan(x)-x-\dfrac{\tan^3(x)} 6 \geq0$. It's derivative is 
$\dfrac{(\tan^2(x))(2-\sec^2x)}{2}$ which is non negative in the mentioned interval.