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Prove that $\int_0^x f^3 \le \left(\int_0^x f\right)^2$

Let $f$ be a differentiable function on $[0,1]$. $f(0)=0$ and $1\ge f'(x)\ge0$. Show that

$\int_0^1f(x)^3dx\le\left(\int_0^1f(x)dx\right)^2$

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marked as duplicate by Marvis, The Chaz 2.0, JavaMan, Paul, Martin Sleziak Oct 23 '12 at 7:02

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1 Answer 1

Set $$G(t) = \bigg(\int_0^t f(x) dx\bigg)^2 - \int_0^t f(x)^3 dx.$$ Then $$G'(t) = 2f(t)\int_0^t f(x) dx - f(t)^3.$$ We would like to know that $G'(t) \geq 0$ on $[0,1]$, as this with the fact that $G(0) = 0$ will prove the inequality. Now, $f(t)$ is obviously nonnegative from the hypothesis given, so it suffices to show that $$H(t) = 2\int_0^t f(x) dx - f(t)^2$$ is nonnegative. We differentiate:

$$H'(t) = 2f(t) - 2f(t)f'(t) = 2f(t)[1 - f'(t)].$$ All of these terms are positive by the hypothesis, so $H'(t) \geq 0$ and $H(0) = 0$ implies $H(t) \geq 0$ for all $t$. Hence the inequality holds.

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