Show that $f(x)=0$ for all $x\geq0$ I have been struggling with this problem..

Q. Let $f(x)$, $x\geq 0$, be a non-negative continuous function, and let $F(x)=\int_0^x f(t) dt$, $x\geq0$. If for some $c>0$, $f(x)\leq cF(x)$ for all $x\geq 0$, then show that $f(x)=0$ for all  $x\geq0$ .

I have tried everything in my ability, but in vain. I get a feeling that this can be solved using Mean Value theorem. Any ideas? Please help!! 

Let $$\phi(t) = e^{-ct} F(t)$$ Then $\phi(0) = 0$, and $\phi(t) \ge 0$ for all $ t \ge 0$.
Furthermore, $$\phi'(t) = e^{-ct}(F'(t) - c F(t)) \le 0$$hence $\phi(t) = \int_0^t \phi'(\tau) d \tau \le 0$, and so $\phi(t) =0 $ for all $t \ge 0$.
If $ϕ(t)=0$  for all $t≥0$ , then $F(t)=0$  for all $t≥0$ . Since $f$  is continuous, $F$  is differentiable and $F ′ =f$ , hence $f=0$. 
