TITLE: Trying a Calculus question
QUESTION [5 upvotes]: One of my pals asked me to look at this question 
Let $f: [0,1] \to \mathbb{R}$ be differentiable. Suppose that $f(0) = 0$
and $0 < f'(x) < 1$ for all $x \in (0, 1)$, where $f'(x)$ is the derivative of $f$. Prove that
$$ \left(\int_{0}^{1}f(x) \ dx \right)^{2} \geq \int_{0}^{1}(f(x))^{3} \ dx$$
I don't really have a clue as to where to start. Any ideas will be appreciated.

REPLY [7 votes]: This Problem is given in the book

Problems in Mathematical Analysis - III by Kaczor and Nowak 

The solution is as follows:
Set $$F(t)= \biggl(\int\limits_{0}^{t} f(x) \ \text{dx}\biggr)^{2} - \int\limits_{0}^{t} (f(x))^{3} \ \text{dx}, \quad t \in [0,1]$$
Then $$F'(t)=f(t) \cdot \biggl(2 \int\limits_{0}^{t} f(x) \ \text{dx} - (f(t))^{2}\biggr)$$ and if $$G(t)= 2 \int\limits_{0}^{t} f(x) \ \text{dx} - (f(t))^{2}$$, then $G'(t)=2f(t) \cdot (1-f'(t)) \geq 0$. Consequently, $G(t) \geq G(0)=0$, which gives $F'(t) \geq 0$. So, $F(t) \geq 0$,, and in particular $F(1) \geq 0$.
Moreover if$F(1)=0$, then $F(t)=0$ for $t \in [0,1]$ and therfore $F'(t)=f(t)G(t)=0$. This, in turn, implies $G'(t)=2f(t) \cdot (1-f'(t))=0$ and $1-f'(t)=0$ for $t \in (0,1)$.