# Prove that $f'(x) > \frac {f(x)}{x}$ for a continuous, differentiable $f(x)$

I'm trying to prove the following:

Say $$f:[0,1] \to \mathbb{R}$$ is continuous on [0,1] and differentiable on (0,1). We also have $$f(0)=0$$ and $$f'$$ is strictly increasing. I want to prove that $$f'(x) > \frac {f(x)}{x}$$.

So far I've tried proof by contradiction, so assume that $$f'(x) \leq \frac {f(x)}{x}$$. I'm guessing this assumption must either contradict $$f$$ being continuous, differentiable, or concave up, however, I can't seem to figure it out. Any pointers would be much appreciated!

• The inequality you want to prove is a rearrangement of $[f(x)/x]'>0$. Maybe that helps? – Arthur Nov 21 '18 at 22:06
• The claim is false for $f(x) = x$. You want to have $f'$ strictly increasing. – daw Nov 21 '18 at 22:12
• @Arthur I did manage to get there by rearranging, but I'm not sure where to go from there. It seems like I need MVT but I'm still having trouble – darcy Nov 21 '18 at 23:18

## 2 Answers

Due to mean value theorem exists $$\xi\in ]0, x[$$ such that $$f(x)=f(x)-0=f(x)-f(0)= xf'(\xi) because $$\xi.

Since $$f'$$ is strictly increasing, you have that $$f'(a)\lt f'(b)$$ for all $$a\lt b$$, which implies that $$f(x)=\int_0^x f'(t)dt\lt \int_0^x f'(x)dt\lt xf'(x)$$ so you have that $$f(x)\lt xf'(x)$$ and you are done.

• That's fine (if $f'$ is strictly increasing), but I'll bet that this exercise appears before the students have encountered integration, as it can be addressed simply with something like the mean value theorem. – John Hughes Nov 21 '18 at 22:19