Arcsin estimation How do I prove that  $$\arcsin (x)>\frac{3}{1+2\sqrt{1-x^2}}\text{ ?}$$
We received this example while we are learning integration so it must have something to do with it. 
But I can't seem to find a function which I should integrate to get this.
 A: I think the original inequality was different and tighter. 

The similar-looking Shafer-Fink inequality states that:
  $$ \forall x\in(0,1),\qquad \arcsin(x)>\frac{3x}{2+\sqrt{1-x^2}}. $$

Proof. If we set $x=\sin\theta$, we have to prove that
$$\forall \theta\in\left(0,\frac{\pi}{2}\right),\qquad  2+\cos(\theta)>3\cdot\frac{\sin(\theta)}{\theta}$$
but
$$ f(\theta)=2+\cos\theta-3\,\frac{\sin\theta}{\theta} = \sum_{n\geq 2} \frac{(-1)^n \theta^{2n}}{(2n)!}\left(1-\frac{3}{2n+1}\right) $$
is an increasing function on $\left[0,\frac{\pi}{2}\right]$ by inspecting the coefficients of the Tayor series on the right.

In the same spirit, we may find a linear combination of $1,\cos(\theta),\cos\left(\frac{\theta}{2}\right)$ whose Taylor series at the origin is the same as the one for $\frac{\sin\theta}{\theta}$ up to the $\theta^4$ term. That leads to the inversion of a Vandermonde matrix and the approximation
$$ 6+7\cos\theta+32\cos\frac{\theta}{2}\approx 45\cdot\frac{\sin\theta}{\theta}$$
that in fact holds as a tight inequality ($\leq$) over $\left[0,\frac{\pi}{2}\right]$. By replacing $\theta$ with $\arcsin x$ and exploiting the cosine bisection formulas we get the improved inequality:

$$ \forall x\in(0,1),\qquad \arcsin(x) < \frac{45x}{6+7\sqrt{1-x^2}+16\sqrt{2}\sqrt{1+\sqrt{1-x^2}}}.$$

A: It's more $$\arcsin(x)<\frac{3}{1+2\sqrt{1-x^2}}.$$
Just show that $\frac{3}{1+2\sqrt{1-x^2}}\geq 1$ and you'll (almost) get the result.
A: It helps to know that 
$$arcsin(x) +c= \int \frac{1}{\sqrt{1-x^2}}dx, c \in \mathbb{R}$$
And $$\forall x \in[-1, 1]:-\frac{\pi}{2} \le arcsin(x) \le \frac{\pi}{2}$$
