Help proving an inequality with arcsin function using Mean value theorem So i have an inequality:
$$ \frac{b-a}{\sqrt{1-a^2}} < \arcsin(b) - \arcsin(a) < \frac{b-a}{\sqrt{1-b^2}}$$
 $$0\leq a < b < 1$$
So I kinda see that both sides look similar to derivative, but how can use it in here?
This is probably going to use Lagrange mean value theorem, or am I mistaken?
I wonder how can I use the fact:
$$\frac{d}{dx}(arcsin x)= \frac{1}{\sqrt{1-x^2}}$$
So I kinda see that $$(b-1)(\arcsin(a))' ?<\arcsin(b)-\arcsin(a)<(b-1)(\arcsin(b))'$$
The right inequality, should work since it's Lagrange theorem usage on it? right?
I wonder that to to the left side, since I put a question mark there.
Is it just the same just because of the fact: $a<b$ that it works the other way around as the right side one?
Any help regarding my solution would be helpful, if it's right or wrong.
 A: You are on the rigth way. Note that by the mean value theorem there exists $c\in (a,b)$ such that 
$$\arcsin b-\arcsin a=\dfrac{b-a}{\sqrt{1-c^2}}.$$ Now, note that
$$0\le a<c<b<1\implies\dfrac{1}{\sqrt{1-a^2}}<\dfrac{1}{\sqrt{1-c^2}}<\dfrac{1}{\sqrt{1-b^2}}$$ and you are done.
A: The easiest way is probably to use monotonicity of the integral:
$$ \frac{1}{\sqrt{1-a^2}} < \frac{1}{\sqrt{1-x^2}} < \frac{1}{\sqrt{1-b^2}}  $$
for $x \in (a,b)$ since $(1-x^2)^{-1/2}$ is decreasing. Integrating,
$$ \frac{b-a}{\sqrt{1-a^2}} < \int_a^b\frac{dx}{\sqrt{1-x^2}} = \arcsin{b}-\arcsin{a} < \frac{b-a}{\sqrt{1-b^2}}. $$
A: Consider $f(x) = \arcsin x$, and pick two values $a$ and $b$ that satisfy $0\le a<b<1$.
By mean value theorem, there exists a $c\in(a,b)$ that satisfies
$$\begin{align*}
\frac{f(b) - f(a) }{b-a} &= f'(c)\\
\frac{\arcsin b - \arcsin a}{b-a} &= \frac{1}{\sqrt{1-c^2}}\\
\end{align*}$$
Within that range $x\in(a,b)$, $f'(x) = \frac1{\sqrt{1-x^2}}$ is strictly increasing,
$$f''(x) = -\frac12\left(1-x^2\right)^{-\frac32}\cdot(-2x) > 0$$
so
$$f'(a) < f'(c) < f'(b)$$
