Use infinite series to prove Use infinite series to prove that
$$\arcsin{x}\lt \frac{x}{1-x^2},$$ for $0\lt x\lt1$.
 A: The first few terms of the Taylor series expansion of $\arcsin(x)$  are: 
$x + \frac{x^3}{6}+ \frac{3x^5}{40}+\frac{5x^7}{112}+\frac{35x^9}{1152}...$
The first few terms of the 
Taylor series expansion of $\frac{x}{1-x^2}$ are: $x + x^3 + x^5 + x^7 ...$
Obviously, the inequality is true.
A: From AM-GM inequality we have, for a given $t$ in the interval $(0,1)$,
$$\frac12\left[\frac{1}{(1+t)^2}+\frac{1}{(1-t)^2}\right]\ge \frac1{(1+t)(1-t)}\tag{1}$$
Now, since $1>\sqrt{(1+t)(1-t)}$, it follows
$$\frac1{(1+t)(1-t)}>\frac{\sqrt{(1+t)(1-t)}}{(1+t)(1-t)}=\frac1{\sqrt{(1+t)(1-t)}}=\frac1{\sqrt{1-t^2}}\tag{2}$$
From $(1)$ and $(2)$ 
$$\frac12\left[\frac{1}{(1+t)^2}+\frac{1}{(1-t)^2}\right]>\frac1{\sqrt{1-t^2}}$$
Let $0<x<1$, integration on the interval $0<t<x$ give us
\begin{align*}
\frac12\int_0^x\left[\frac{1}{(1+t)^2}+\frac{1}{(1-t)^2}\right]dt&>\int_0^x\frac1{\sqrt{1-t^2}}dt\\
\frac12\left[-\frac{1}{1+x}+\frac1{1-x}+1-1\right]&>\arcsin x-0\\
\frac{x}{1-x^2}&>\arcsin x
\end{align*}
A: Let $f(x)=\frac{x}{1-x^2}-\arcsin{x}$.
Hence, $$f'(x)=\frac{1-x^2-x(-2x)}{(1-x^2)^2}-\frac{1}{\sqrt{1-x^2}}=\frac{1+x^2-\sqrt{(1-x^2)^3}}{(1-x^2)^2}=$$
$$=\frac{ x^2(x^4-2x^2+5)}{(1+x^2-\sqrt{(1-x^2)^3})(1-x^2)^2}>0.$$
Thus, $f(x)>f(0)=0$ and we are done!
