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Proving Integral Inequality

Suppose $f(x)$ is differentiable on $[0,1]$ , $f(0)=0$ and $1\geq f'(x) >0 $

Prove that $\displaystyle\left(\int_{0}^{1} f(x)\;dx\right)^2\geq\int_{0}^{1}\left(f(x)\right)^3\;dx$

Sorry about that , I have no idea to start the prove

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marked as duplicate by JavaMan, Marvis, Douglas S. Stones, Martin Argerami, Mhenni Benghorbal Jan 15 '13 at 5:56

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Let $F(t)=\displaystyle\left(\int_{0}^{t} f(x)dx\right)^2-\int_{0}^{t}\left(f(x)\right)^3dx$, then $F'(t)=f(t) \left(2\int_{0}^{t} f(x)dx-(f(t))^2\right).$

Let $G(t)=2\int_{0}^{t} f(x)dx-(f(t))^2$,then $G'(t)=2f(t)[1-f'(t)]\geq 0,$ so $G(t)\geq G(0)=0.$

And this implies that $F'(t)\geq0.$ Hence:$$0=F(0)\leq F(1)=\displaystyle\left(\int_{0}^{1} f(x)dx\right)^2-\int_{0}^{1}\left(f(x)\right)^3dx.$$ It's over!

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Hint: Consider $$F(t) = \left(\int_0^t f(x)\, dx \right)^2 - \int_0^t \left(f(x)\right)^3\, dx$$

Consider $F'(t) = f(t) G(t)$. Show that $f(t) \geq 0$. Consider $G'(t)$, show that $G(t) \geq 0$.

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