Prove $f(x)\le0$ if $f(0)=0$ and $\int_0^xf(t)\mathbb dt\ge xf(x)$ 
$f(x)$ 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 $x$.

Hint: write is as $$\int_0^x \big(f(t)-f(x)\big) dt \ge 0$$

[ EDIT ]   The above implies that for a given $x$, there exists a $c \in (0,x)$ such that $f(c) \ge f(x)$. Therefore the set $C_1 = \{ c \in (0,x) \mid f(c) \ge f(x)\}$ is not empty and, being obviously bounded, it has an infimum $0 \le c_1 = \inf C_1 \lt x$. Also, it follows by continuity that $f(c_1) \ge f(x)$.

Repeat for interval $(0,c_1)$ and so on, building the decreasing sequence $x \gt c_1 \gt c_2 \gt \cdots$ with $f(x) \le f(c_1) \le f(c_2) \le \cdots$. Let $c = \lim_{n \to \infty} c_n$, then again by continuity $f(x) \le f(c)$.
If $c=0$ then $f(x) \le f(0) = 0$ and the proof is complete. Otherwise $c \gt 0$ and it follows fairly easily that $f(t) \le f(x)$ for $t \in (0,c)$. If $f$ is constant on $(0,c)$ then yet again again by continuity $f(x) \le f(c) = f(0) = 0$. Otherwise there must exist a $t_0 \in (0,c)$ where strictly $f(t_0) \lt f(c)$. But then the strict inequality would hold on a neighborhood of $t_0$ which, together with $f(t) \le f(c)$, would cause $\int_0^c \big(f(t)-f(c)\big) dt \lt 0$ contradicting the premise.
