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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!

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  • $\begingroup$ The inequality you want to prove is a rearrangement of $[f(x)/x]'>0$. Maybe that helps? $\endgroup$ – Arthur Nov 21 '18 at 22:06
  • $\begingroup$ The claim is false for $f(x) = x$. You want to have $f'$ strictly increasing. $\endgroup$ – daw Nov 21 '18 at 22:12
  • $\begingroup$ @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 $\endgroup$ – darcy Nov 21 '18 at 23:18
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Due to mean value theorem exists $\xi\in ]0, x[$ such that $$ f(x)=f(x)-0=f(x)-f(0)= xf'(\xi)<xf'(x) $$ because $\xi<x$.

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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.

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  • $\begingroup$ 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. $\endgroup$ – John Hughes Nov 21 '18 at 22:19

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