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 $$g(x)=\sin x-(x-\frac{x^3}{6})=\frac{x^3}{6}-x+\sin x$$. One has 

$$g'(x)=\frac{x^2}{2}-1+\cos x$$ $$g''(x)=x-\sin x$$ $$g'''(x)=1+\cos x$$
It is clear that $g'''(x)\ge 0$ so $g''(x)$ is increasing so positive because $g''(0)=0$ which implies $g'(x)$ is increasing so positive since $g'(0)=0$. But this implies that $g(x)$ is increasing and positive because $g(0)=0$ 

Thus $$\sin x>x-\frac{x^3}{6}$$ (all on $0<x<\pi$)