Q:

Integral inequality problem

Let $f:[0,1]\to\mathbb R$ be a differentiable function with $f(0)=0$ and $f'(x)\in(0,1)$ for every $x\in(0,1).$  Show that
$$\left(\int_0^1f(x)dx\right)^2>\int_0^1(f(x))^3dx$$
I am not even sure how to begin to solve this problem.  Any help/hints is appreciated.  Thanks.

A:

Denote 
$$
F(t)=\left(\int_0^tf(x)dx\right)^2-\int_0^t(f(x))^3dx, t\in[0,1]
$$
Then we have
$$
F'(t)=f(t)\left(2\int_0^tf(x)dx-(f(t))^2\right)
$$
and if $G(t)=2\int_0^tf(x)dx-(f(t))^2$ then $G'(t)=2f(t)(1-f'(t))\ge 0$.
Then, $G(t)\ge G(0)=0$ giving $F'(t)\ge 0$. Thus $F(t)\ge 0$, in particular $F(1)\ge 0$.

