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

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 $x>0$.