I found that I could do this with the Mean Value theorem and also found a similar question here in the stackexchange and Socratic but couldn't solve the right hand side of the inequality.
I figured that by specifically choosing the function and the interval (combinations of the given question) we could prove that. But I am lost tin selecting the appropriate function.
Can you guys please suggest some ideas and also some example sums where I could check these kind of sums.
Thanks
Let $f(x)=x-\sin^{-1} x \implies f'(x)=1-\frac{1}{\sqrt{1-x^2}}=\frac{\sqrt{1-x^2}-1}{\sqrt{1-x^2}}<0$ So $f(x)$ is a decreasing function. $$x\ge 0 \implies f(x)\le f(0) \implies f(x)\le 0$$ Next, we take $g(x)=\sin^{-1} x-\frac{x}{\sqrt{1-x^2}} \implies g'(x)=\frac{-x^2}{\sqrt{1-x^2}}<0.$ So $g(x)$ is decreasing function, then $$ x \ge 0 \implies g(x)\le g(0)=0 \implies \sin^{-1}x \le \frac{x}{\sqrt{1-x^2}}.$$
For $|x|<1$,
$$1<\frac1{\sqrt{1-x^2}}<\frac1{(1-x^2)^{3/2}}$$ seems obvious. Then by integration between $0$ and $x$ you get the claimed inequalities.
This is pretty standard property of $\arcsin$ which is equivalent to the more famous inequality $\sin x <x<\tan x$ for $x\in(0, \pi/2)$.
A direct proof can be given by noting that if $x\in(0,1)$ and $t\in(0,x)$ then $$1<\frac{1}{\sqrt{1-t^2}}<\frac{1}{\sqrt{1-x^2}}$$ integrating the above with respect to $t$ on interval $[0,x]$ we get $$x<\arcsin x<\frac{x} {\sqrt{1-x^2}}$$ for all $x\in(0,1)$.
You can avoid integrals and achieve the same result via mean value theorem. If $x\in(0,1)$ then by mean value theorem we have $$\arcsin x=\arcsin x - \arcsin 0=\frac {x}{\sqrt{1-c^2}}$$ for some $c\in(0,x)$. But then we can obviously see that $$1<\frac{1}{\sqrt{1-c^2}}<\frac{1}{\sqrt{1-x^2}}$$ Multiplying by positive $x$ we get $$x<\arcsin x<\frac{x} {\sqrt{1-x^2}}$$
