Cfrgtkky

Define $f : [0,1] to mathbb{R} $ differentiable satisfying $f(0)=0$ and $0<f'(x)<1$.

Prove: $displaystyleint_{0}^{1} f^3(x) , dx < left(displaystyleint_{0}^{1} f(x) , dxright)^2$

I found the following answer,

Define $F(x)=displaystyleint_{0}^{x} f^3(t) , dt - left(displaystyleint_{0}^{x} f(t) , dtright)^2$

We have $$F'(x)=f^3(x)-2 left(displaystyleint_{0}^{x} f(t) , dtright)f(x)=f(x)left[f^2(x)-2 left(displaystyleint_{0}^{x} f(t) , dtright)right]$$
and
$$left[f^2(x)-2 left(displaystyleint_{0}^{x} f(t) , dtright)right]'=2f(x)left[f'(x)-1right]$$
from which we can arrive at the conclusion.

But I felt a little confused.

In the past I usually proved the inequality by classical inequalities and something like Taylor expansion to use the condition about derivative.

But this answer is little out of my expectation.

Do we have other proofs or ideas to arrive at $displaystyleint_{0}^{1} f^3(x) , dx < left(displaystyleint_{0}^{1} f(x) , dxright)^2$? And other thoughts about solving these types of issues?

I think when we talk about $left(displaystyleint_{0}^{1} f(x) , dxright)^2 >ldots ,$ it's a little difficult to deal with. I don't quite know what tools to use.

Clearly $f(x) > 0$ for all $x in langle 0,1]$. As the solution states, we have
$$left[f(x)^2-2 left(displaystyleint_{0}^{x} f(t) , dtright)right]'=2f(x)left[f'(x)-1right] < 0$$
so $f(x)^2 < 2displaystyleint_0^xf(t),dt$ for all $x in langle 0,1]$. Also notice that $left(displaystyleint_{0}^{x} f(t) , dtright)' = f(x)$ so

begin{align}
int_0^1f(x)^3,dx &= int_0^1f(x)^2, f(x),dx \
&< int_0^12 left(displaystyleint_{0}^{x} f(t) , dtright)left(displaystyleint_{0}^{x} f(t) , dtright)',dx \
&= left(displaystyleint_{0}^{x} f(t) , dtright)^2Bigg|_0^1 \
&= left(displaystyleint_{0}^{1} f(t) , dtright)^2
end{align}

$$f(x)^2 = int_0^x 2f(t),f'(t),dt< 2displaystyleint_0^xf(t),dt$$

Write $F(x) = int_0^x f(x)dx$ and $I= int_0^1 f(x)dx$. Using integration by parts formula, we get
$$begin{eqnarray}
int_0^1 f(x)^3dx &=& (F(x)-F(1))f(x)^2 Big|^{x=1}_{x=0} + 2int_0^1 (F(1)-F(x))f(x)f'(x)dx\
&<&2int_0^1 (F(1)-F(x))f(x)dx = -(F(1)-F(x))^2Big|^{x=1}_{x=0} = F(1)^2 =I^2.
end{eqnarray}$$

Note: However, this calculation is only valid under the assumption that $f$ is absolutely continuous, or $f'$ is Riemann integrable.