TITLE: Prove $f(x)\le0$ if $f(0)=0$ and $\int_0^xf(t)\mathbb dt\ge xf(x)$
QUESTION [6 upvotes]: $f(x)$ is a differentiable real valued function that satisfies following conditions:
  $$f(0)=0$$
  $$\int_0^xf(t)\mathbb dt\ge xf(x)\quad$$
  Prove that for all $x>0$
  $$f(x)\le0$$

I tried but couldn't derive the conclusion from the given conditions. $f(x)$ seems to have to decrease around $x=0$ but not necessarily for all $x$.

REPLY [5 votes]: Hint. Consider the function $F(x):=\frac{1}{x}\int_0^x f(t)dt$ and show that:
i) $\lim_{x\to 0^+} F(x)=f(0)=0$.
ii) $F(x)\geq f(x)$ for $x>0$.
iii) $F'(x)\leq 0$ for $x>0$.
Then $F(x)$ is decreasing and $0\geq F(x)\geq f(x)$ for $x>0$.