Show $ f(x) < x\cdot f'(x)$ Function $f:[0,\infty) \to \mathbb R$ continuous on $[0,\infty)$ and differentiable at $(0,\infty)$. 
$f(0)=0$.
$f'$ is strictly increasing.

You can apply Lagrange's theorem (mean value theorem):
$$
\frac{f(x)-f(0)}{x-0}=f'(c)
$$
for some $c\in(0,x)$. Since $f(0)=0$ and $x>0$, we get
$$
f(x)=xf'(c)
$$
Now apply the fact that $f'$ is strictly increasing.
