 Solution

The assumptions imply $f\ge 0$. Since $(f^2)\prime =2ff\prime \le 2f$ and both sides vanish at zero, integration gives $f(x{)}^2\le 2F(x)$ where $F(x)={\int }_0^{x}f$. NowThe bracket vanishes at zero, proving ${\int }_0^x{f}^3\le ({\int }_0^{x}f{)}^2$.