I need to show that if $f$ is a $C^1$ function with $f( 0 ) = 0$ and $f'( x ) > f( x )$ for all $x \in \mathbb{ R }$ then $f( x ) > 0$ for all $x > 0$.
I think I need to show that $f( x ) < 0$ for some $x$ leads to a contradiction ( $f( x ) \neq 0$ for some $x \neq 0$ follows ). I know that if there is at least one point $x_0$ such that $f( x_0 ) < 0$ then by the mean value theorem, I can find infinitely many. So $f \to \infty$ on $( 0, x_0 )$.
I'm not sure how I can exactly reach the contradiction from this knowledge.