How to show that $f(x)=\frac{\arcsin x}x$ is increasing when $x\ge0$? How to show that $f(x)=\frac{\arcsin x}x$ is increasing when $x\ge0$?
My Attempt:
$f'(x)=\frac{\frac x{\sqrt{1-x^2}}-\arcsin x}{x^2}$
Since $0\le x\le1\implies0\le\sqrt{1-x^2}\le1$
And $0\le\arcsin x\le1.57$
But not able to show that $f'(x)\ge0$

You can also find a series expansion of $\arcsin(x)$ as follows:
$$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\dots$$
Put $x=\frac{t^2}{1-t^2}$, and we have
$$\frac{1}{\sqrt{1-t^2}}=1+\frac{t^2}{2}+\frac{3t^4}{8}+\frac{5t^6}{16}+\dots$$
Now integrating we have
$$\arcsin(x)=\int_0^x\frac{1}{\sqrt{1-t^2}}dt=x+\frac{x^3}{6}+\frac{3x^5}{40}+\frac{5x^7}{112}+\dots$$
and so $$\frac{\arcsin(x)}{x}=1+\frac{x^2}{6}+\frac{3x^4}{40}+\frac{5x^6}{117}+\dots$$
which is clearly increasing for $x\geq 0$.
