Prove: $\sin (\tan x) \geq {x}$ 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.. 
 A: 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.
A: Let $t = \tan x$. Then you need to prove $\sin t \geq \arctan t$ for $t \in [0,1]$.
If you look at $f(t) = \sin t - \arctan t$, we have $f'(t) = \cos t - \frac{1}{1 + t^2}$. But the inequalities
$$\frac{1}{1 + t^2} \leq 1 - \frac{1}{2}t^2 \leq \cos t, \quad (0 \leq t \leq 1)$$
are straightforward, showing that $f'(t) \geq 0$. Therefore $f$ is increasing on $[0,1]$. Because $f(0) = 0$, it follows that $f(t) \geq 0$, which is the desired inequality.
To prove the inequality $\cos t \geq 1 - \frac{1}{2}t^2$, let $g(t) = \cos t - 1 + \frac{1}{2}t^2$, and show that $g(0) = 0$, $g'(t) \geq 0$ for $t \geq 0$.
A: 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 $x \in \Bigl[0,\dfrac{\pi}{4}\Bigr]$.
A: Another approach which is much less satisfactory than what has been given in the previous answers would consist in building the Taylor expansion of $\sin(\tan(x))$ around $x=0$.   
This leads to $x + \dfrac {x^3}{ 6} - \dfrac { x^5 }{ 40} - \dfrac {55 x^7} { 1008} - \dfrac {143 x^9 }{ 3456} + \cdots$  
Then $\sin(\tan(x)) - x = \dfrac {x^3}{ 6} - \dfrac { x^5 }{ 40} - \dfrac {55 x^7} { 1008} - \dfrac {143 x^9 }{ 3456} + \cdots$ which is an increasing function of x. For $x=\dfrac{\pi}{4}$, the values of the different terms are $0.0807455, 0.0074712, 0.0100585, 0.0047051$.
