Let $f:[0,1]\to \Bbb R$ be differentiable function with $f(0)=0$ and $0\leqslant f(x)'\leqslant 1$. Then $$3\left(\int_0^1f(x)^2dx\right)^3\leqslant \int_0^1f(x)^8dx.$$
My attempt:
$0\leqslant f(x)'\leqslant 1$ implies $0\leqslant f(x)\leqslant x$. then $0\leqslant f(x)^2\leqslant xf(x)$ and $0\leqslant f(x)^2\leqslant x^2$.
$$0\leqslant \int_0^1f(x)^2\leqslant 1/3$$
But these inequality can't relate $f(x)^2$ and $f(x)^8$. Any idea?
Your inequality does not seem to be true in general. Take $f(x)=ax$ with $a<1$: for this function we have that $0<f'<1$ and $f(0)=0$, so it is a legit function according to your requests. By using your inequality, a direct calculation gives $(a^6)/9<(a^8)/9$, namely $1<a^2$. This is a contradiction since we assumed $a<1$.