Integral inequality $\int_0^x{f(t)^3 dt \leq \left( \int_0^x f(t) dt\right)^2} :\forall x>0$ Let $f(0) = 0$ and $0<f'(x)\leq1$ for all $x \geq0$, then prove:
$$\int_0^x{f(t)^3 dt \leq \left( \int_0^x f(t) dt\right)^2} :\forall x>0$$
The hint I was given was "differentiate, factor and differentiate again" but I'm not sure where to start

Let $g(x)=\left(\int_0^xf(t)dt\right)^2-\int_0^xf(t)^3dt$. We want to show $g(x)\geq0$ for all $x\geq0$. Clearly $g(0)=0$, so it suffices to show $g'(x)\geq0$ for all $x\geq0$. Differentiating, we get
$$g'(x)=2f(x)\int_0^xf(t)dt-f(x)^3=f(x)\left[2\int_0^xf(t)dt-f(x)^2\right].$$
Since $f(0)=0$ and $f'(x)>0$ for all $x\geq0$, $f(x)\geq0$ for all $x\geq0$. So all that is left is to show $h(x)=2\int_0^xf(t)dt-f(x)^2\geq0$ for all $x\geq0$, which you can do in more or less the same way.
