Proving that $x-\frac{x^{3}}{6} < \sin(x)$ for $0I have to prove that $$x-\frac{x^{3}}{6} < \sin(x)  \quad\text{ for }\quad  0<x<\pi $$
I tried to define $f(x) = \sin(x) - (x-\frac{x^{3}}{6})$, and to differentiate it,
but $f'(x) = \cos(x) -1 +\frac{1}{2}x^{2}$. I have no idea how to continue.
Thanks.

Consider that
$$
\sin(x)=\sum_{n\ge 0} (-1)^n\frac{x^{2n+1}}{(2n+1)!}.
$$
Moreover, for all $x \in [0,\pi]$ and positive $n$ it holds $ (2n+3)(2n+2)\ge 20>\pi^2\ge x^2$, so that
$$
\left|\frac{x^{2n+1}}{(2n+1)!}\right|> \left|\frac{x^{2n+3}}{(2n+3)!}\right|.
$$
Therefore for each integer $m\ge 0$ and $x \in [0,\pi]$ we obtain
$$
\sum_{n=0}^{2m+1} (-1)^n\frac{x^{2n+1}}{(2n+1)!} <\sin(x)<\sum_{n=0}^{2m} (-1)^n\frac{x^{2n+1}}{(2n+1)!}.
$$
In particular, we have
$$
x-\frac{x^3}{6}<\sin(x)<x-\frac{x^3}{6}+\frac{x^5}{120}.
$$
