Proof of $\arcsin x \le 2\arctan x$? I am looking for a proof for the following 'fact':
$$
\arcsin x \le 2\arctan x \quad \forall x\in[0,1).
$$
I put fact between single quotes, as the only proof I found is a plot by wolframalpha. I know the relationship between $\arcsin$ and $\arctan$,
$$
\arcsin(x) = \arctan\left( \frac x{\sqrt{1-x^2}}\right), \ x\in[0,1),
$$
however, I do not know how to take advantage of this. (Of course, this gives a lower bound, $\arcsin x\ge \arctan x$, which is not what I am looking for)
 A: As $x\in[0,1);0<\arcsin x,2\arctan x<\dfrac\pi2$ where sine is increasing 
Method $\#1:$
$\sin(\arcsin x)=x$
Setting $\arctan x=y\implies \tan y=x,$
We have, $\sin(2\arctan x)=\sin2y=\dfrac{2\tan y}{1+\tan^2y}=\dfrac{2x}{1+x^2}$
Now, $x\le\dfrac{2x}{1+x^2}\iff x(x^2-1)\le0$
The equality holds if $x=0$
and for $x>0,$ we need $x^2<1\iff-1\le x\le1$ and we have $0\le x<1$
Method $\#2:$
Using the definition of principal values, here $0\le2\arctan x<\dfrac\pi2$
Like the other method,  $\sin(2\arctan x)=\dfrac{2x}{1+x^2}\implies2\arctan x=\arcsin\left(\dfrac{2x}{1+x^2}\right)$
Now as $\arcsin$ is increasing in $[0,1)$ actually in $[-1,1];$
it is sufficient to establish $x\le\dfrac{2x}{1+x^2}$
A: It is just a convexity inequality. Both the functions have the same values in the endpoints of $[0,1]$, but $\arcsin x$ is a convex increasing function (since the inverse function $\sin x$ is concave over $(0,\pi/2)$) while $2\arctan x$ is a concave increasing function (since $\tan x$ is convex over $(0,\pi/2)$). 
It follows that, over $(0,1)$:
$$\arcsin x < \frac{\pi x}{2} < 2\arctan x. $$
A: Hint Use the addition formula of inverse tangent function
$$\arctan x+\arctan y=\arctan\frac{x+y}{1-xy}$$
and use the fact that $\arctan$ is increasing function.
A: use that $f(x)=2\arctan(x)-\arcsin(x)$ have a negative second derivative, namely
$f''(x)=-\frac{x}{\left(1-x^2\right)^{3/2}}-\frac{4 x}{\left(x^2+1\right)^2}$
A: Denote $f(x)=\arcsin x-2\arctan x$, then $f''(x)\geq0$ for $x\in[0,1)$, so
$$f(x)\leq(1-x)f(0)+xf(1)=0$$
