Question: I 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.

Answer: 
In fact $x-x^3/6< \sin x$ for all $x > 0.$ Proof: As you noted, we can define $f(x) = \sin x -x + x^3/6.$ Is $f(x)>0$ for $x > 0?$ Because $f(0)=0,$ this is true if $f'(x) = \cos x -1 + x^2/2 > 0$ for $x>0.$ Is this true? Because $f'(0)=0,$ this will hold if $f''(x) = -\sin x + x > 0$ for $x> 0.$ Well, hopefully the last inequality looks familiar.
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
$$
x-\frac{x^3}{6}<\sin(x)<x-\frac{x^3}{6}+\frac{x^5}{120}.
$$