proving an integral inequality_ I'm having problem proving the following integral inequality.
Let $f$ be a continuous function on $[0,1]$ such that $f(0)=0$ and $f'(x)\geq1$ for all $x\in(0,1)$. Show the following inequailties hold.
$$\left( \int^{1}_{0} f(x) dx \right)^2 \leq \int ^{1} _{0} (f(x))^3 dx$$
I have tried using C-S inequalities of integrals, and also integration by parts, but failed. Please help thank you!
 A: As in the previous post, set $g(x) = f(x) - x$. The desired inequality is equivalent to
$$
\int_0^1 g(x) dx + \left(\int_0^1 g(x) dx \right)^2 \le \int_0^1 3x^2 g(x) dx + \int_0^1 3x g^2(x) dx + \int_0^1 g^3(x) dx \, .
$$
Now observe that for any $h:[0,1]\to \mathbb{R}$ with $h(0) = 0$ and $h' \ge 0$ the expression
$$
\alpha \mapsto \int_0^1 \alpha x^{\alpha - 1} h(x) dx = h(1) - \int_0^1 x^\alpha h'(x) dx
$$
is increasing in $\alpha \in [1, \infty)$. In particular
$$
\int_0^1 h(x) dx \le \int_0^1 2x h(x) dx \le \int_0^1 3x^2  h(x) dx
$$
for all such $h$. Applying this with $h = g$ we obtain
$\int_0^1 g(x) dx \le \int_0^1 3x^2 g(x) dx$. Using Cauchy-Schwarz and applying this estimate with $h = g^2$ we obtain
$$
 \left(\int_0^1 g(x) dx \right)^2 \le  \int_0^1 g^2(x) dx \le  \int_0^1 2x g^2(x) dx \, .
$$
Combining all this the desired estimate follows.
A: Given ,
$ g(x)=f(x)-x$ and $ g(0)=0 $ , $ g'(x)\ge 0$
So,
$$\begin{equation}\begin{aligned}
 \int_0^1 f^3(x)dx - \left(\int_0^1 f(x)dx\right)^2 &=\int_0^1 (g(x)+x)^3dx - \left(\int_0^1 g(x)+xdx\right)^2 \\
  &=\int_0^1 g^3(x)+3xg^2(x)+3x^2g(x)+x^3 dx -\left(\int_0^1 g(x)dx +\frac{1}{2}\right)^2 \\
  &=\int_0^1 g^3(x)+3xg^2(x)+(3x^2-1)g(x)dx -\left(\int_0^1 g(x)dx\right)^2 \\
  &\ge \int_0^1 g^3(x)dx +\int_0^1(3x-1)g^2(x)dx+\int_0^1(3x^2-1)g(x)dx\\
  &\ge \frac{1}{2}g^2(\frac{1}{3})\\
  &\ge 0
\end{aligned}\end{equation}$$
A: Firstly notice that $f(x)$ is increasing, so $f(x) \geq 0$
Let $F(x) = \int_0^x f(t)dt$. Then $f(x)^2 - 2F(x)$ is increasing ($(f^2(x) - 2F(x))' =2f(x)f'(x) - 2f(x) = 2f(x)(f'(x) - 1) \geq0$).
Then $f^2(x) - 2F(x) \geq f^2(0) - 2F(0) = 0 \Rightarrow (\int_0^xf^3(t)dt - F(x)^2)' = f^3(x) - 2F(x)f(x) =f(x)(f^2(x) - 2F(x)) \geq0$
(where we have apply FTC)
Then $\int_0^xf^3(t)dt - F(x)^2 \geq \int_0^0f^3(t)dt - F^2(0) = 0$. For your inequality, apply at $x=1$, getting
$F(1)^2 \leq \int_0^1 f^3(t)dt \Rightarrow (\int_0^1 f(t)dt)^2 \leq \int_0^1f^3(t)dt$
Equality holds iff $f(x)(f^2(x) - 2F(x))=0 \ \forall x \in (0,1)$. Let $s = sup\{x\in [0,1] | f(x)=0\}$. Then, $s = 0$ (if not, by continuity $f(s) = 0$ and by monocity $F(x) = f(x) =0 \ \forall x \in[0,s]$. But then $f'(x) = 0 \ \forall x \in (0, s)$, contradiction with the assumptions).
Then $2F(x) = f^2(x) \ \forall x \in [0, 1]$. Differenciating, we get $2f(x)(f'(x) -1) = 0 \forall x \in (0,1)$. But $f(x) \geq 0$, so $f'(x) = 1 \forall x \in (0,1)$. Finally $(f(x) - x)' = 0 \ \forall x\in(0,1)$, so $f(x) - x$ is constant and $f(x) - x = f(0) - 0 = 0 \qquad \square$
