If an absolutely continuous function satisfies $f(0)=0$, then $\int_0^x|f(t)f'(t)|dt\leq\frac{1}{2}\big(\int_0^x|f'(t)|dt\big)^2$ 
Assume that $f$ is absolutely continuous on $[0,a]$ for every
    $a\geq0$, and $f(0)=0$. Prove that
    \begin{equation*}
    \int_0^x|f(t)f'(t)|dt\leq\frac{1}{2}\biggl(\int_0^x|f'(t)|dt\biggr)^2
  \end{equation*}

I could show that the inequality holds without the $\frac{1}{2}$, but I have no idea where the half should come from.
Any help?

Since $|f(t)|\le \int_0^t |f'(s)|\,ds$, the inequality follows from the identity
$$
\iint_{0\le s\le t\le x}|f'(t)f'(s)|\,dt\,ds  = \frac 12
\iint_{0\le s,t\le x}|f'(t)f'(s)|\,dt\,ds$$
which holds by virtue of symmetry.
