Let $f:\mathbb{R} \rightarrow \mathbb{R}$ a $C^1$ class function with $0 < f'(t) \leq 1$ for all $t \in [0,1]$ and $f(0)=0$ prove that $$\left(\int_{0}^{1} f(t) dt\right)^2 \geq \int_{0}^{1} f^3(t) dt, $$ where $f^3=f\circ f\circ f$.

My attempt: So I tried to use Cauchy-Schwarz, Integration by Parts, Mean value Theorem for Integrals and wasn't unable to connected the dots. Then after a while I think if I use the identity function I will have a contradition. So now I don't know if this statement is right or wrong. I appreciate any help. Thank you

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    $\begingroup$ With the identity function the equality holds... It is $1/4$ in both sides. $\endgroup$ – DiegoMath Jun 12 '18 at 18:00
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    $\begingroup$ in this case the problem is wrong. But is interesting think in the problem with cubic power function. $\endgroup$ – DiegoMath Jun 12 '18 at 18:09

I think the question was not about function composition because then $f(x)=x$ is a counterexaple. It does hold when you assume it is a third power however:

Notice that $f$ is increasing in $[0,1]$ (since $f'>0$) and $f(0)=0$ thus $f(x)\geq{}0,\forall{}x\in{}[0,1]$. Set $F(x)=\int_0^xf(t)dt$ then if $h(x)=F^2(x)-\int_0^xf^3(t)dt$ we have $h'(x)=2f(x)F(x)-f^3(x)=f(x)(2F(x)-f^2(x))$. Now set $g(x)=2F(x)-f^2(x)$, then $g(0)=0$ and $g'(x)=2f(x)-2f(x)f'(x)=2f(x)(1-f'(x))\geq{}0,\forall{}x\in{}[0,1]$ and thus $g(x)\geq{}0,\forall{}x\in{}[0,1]$. Now $h'(x)=f(x)g(x)\geq{0},\forall{}x\in{}[0,1]$ and thus since $h(0)=0$ and $h$ is increasing we have $h(1)\geq{}h(0)=0$.


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