Integral inequality 4 Let $f$ be a function having a continuous derivative on $[0,1]$and with the property that $0<f'(x)\leq 1, f(0)=0$. I got stuck to prove: 
$$\Bigg[\int_{0}^{1}f(x)dx \Bigg]^2\geq \int_{0}^{1}[f(x)]^3 dx$$
Can someone help me pls? 
 A: We consider the differentiable function $G$ with $G(u)=(\int_{0}^{u}f(x)dx)^2-\int_{0}^{u}f(x)^3dx$ for $u\in [0,1]$.
It is $\ G'(u)=f(u)(2\int_{0}^{u}f(x)dx-f(u)^2),\ \forall u\in [0,1]$.
Now consider the differentiable function $F$ with $F(u)=2\int_{0}^{u}f(x)dx-f(u)^2$ for $u\in [0,1]$. Then:
$F'(u)=2f(u)(1-f'(u)),\ \forall u\in [0,1]$.
We have $f'(u)>0,\ \forall u \in[0,1]$, which means that $f$ is strictly increasing in $[0,1]$ and hence $f(u)\geq f(0)=0\ (1)$, when $u \in [0,1]$. It is also given that $f'(u)\leq 1,\ \forall u\in [0,1]$. Therefore
$F'(u)\geq 0,$ i.e. $F$ is increasing and so $F(u)\geq F(0)=0\ (2)$.
Since $G'(u)=f(u)F(u)$, $(1)$ and $(2)$ imply that $G'(u)\geq 0,\ \forall u \in [0,1]$, i.e. $G$ is increasing and so $G(1)\geq G(0)=0$. This proves our inequality. 
A: From $f(x)-f(0)=f'(\zeta)(x-0)$ for $0<\zeta<1$, and $0<f'(\zeta)\leq1$ so
$$0\leq f(x)\leq1$$
This shows $f^3\leq f^2$ and $\displaystyle\int_0^1f^3\leq\int_0^1 f^2$. In the other hand from inequality for integrals we have $\displaystyle\int_0^1 ff\leq\int_0^1 f\int_0^1 f$ Then
$$\int_0^1f^3\leq\Big(\int_0^1 f\Big)^2$$
