Let $f\in C^1[0,1]$ be a function such that $f(0)=0$ and $0\leq f'(x)\leq 1$ for all $0\leq x\leq 1$. I want to prove that $$ \int_0^1 [f(x)]^3 \; dx \leq \left( \int_0^1 f(x)\; dx \right)^2. $$ I tried by defining $$ g(x) = \int_0^x [f(t)]^3\; dt \quad\text{and}\quad h(x)=\left( \int_0^x f(t)\; dt \right)^2, $$ then the desired inequality becomes $g(1)\leq h(1)$. Computing the first and second derivatives of $g$ and $h$ we obtain $$ g'(x) = [f(x)]^3, \quad h'(x) = 2 f(x) \int_0^x f(t) \; dt $$ and $$ g''(x) = 3[f(x)]^2 f'(x), \quad h''(x) = 2[f(x)]^2 + 2f'(x)\int_0^x f(t)\; dt. $$ As $f(0)=0$, we obtain that $g'(0)=0$, $h'(0)=0$, $g(0)=0$ and $h(0)=0$, and as $0\leq f'(x)\leq 1$ for all $x$, we conclude that $0\leq f(x)\leq x$.
From this it follows that $$ g''(x)/3\leq [f(x)]^2, \quad h''(x)/2\geq [f(x)]^2 $$ and all I was able to conclude from the above (integrating twice) is that $g(x)\leq 3h(x)/2$ for all $x$, but this is far from the goal, so this is not a good approach. Is there any way to prove that $g(1)\leq h(1)$?
Note that \begin{align}\left(\int_{0}^{1} f(x) d x\right)^{2}-\int_{0}^{1} f^{3}(x)\, d x&= 2\int_0^1 f(u)\left(\int_0^u f(v)\,dv\right) du\\ &\qquad\qquad- \int_{0}^{1}f(u)\left(\int_0^u D(f^2(v))\,dv\right)du \\ &=2 \int_{0}^{1}f(u)\left(\int_{0}^{u} f(v)\left(1-f'(v)\right) d v\right)d u \end{align} and the RHS is $\geq 0$ by the given assumptions.