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Suppose $f(x)$ is a nonnegative convex function in $[0,1]$. Prove:

$$\displaystyle \int_0^1f^2(x)\,\mathrm dx\leqslant\frac43\left(\int_0^1f(x)\,\mathrm dx\right)^2$$

I have tried Cauchy Mean Value Theorem: Construct $\displaystyle F(x)=\frac{\displaystyle \int_0^1 f^2(x)\,\mathrm dx}{\displaystyle \left(\int_0^1f(x)\,\mathrm dx\right)^2}$... But it doesn't work :-(
Any tips would be appreciated!

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  • $\begingroup$ Where $f^2(x)\equiv f(f(x))$ or $[f(x)]^2$? $\endgroup$ – Kenny Lau Apr 24 '16 at 2:36
  • $\begingroup$ @KennyLau The second one, most likely. $\endgroup$ – user228113 Apr 24 '16 at 2:36
  • $\begingroup$ @KennyLau $\big(f(x)\big)^2\triangleq f^2(x)$ $\endgroup$ – Shine Mic Apr 24 '16 at 2:37
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    $\begingroup$ Not sure if it will work, but you're in a probability space and you have a convex function, so try Jensen's inequality. $\endgroup$ – Rick Sanchez Apr 24 '16 at 4:03
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Let $f(x)=x^2$. Then $$ \int_0^1 f^2(x)\,dx=\int_0^1x^4\,dx=0.2, $$ while $$ \frac43\,\left(\int_0^1 f(x)\,dx\right)^2=\frac43\,\left(\int_0^1x^2\,dx\right)^2=\frac4{27}=0.\overline{148}. $$

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  • $\begingroup$ All to easy! +1 $\endgroup$ – Mark Viola Apr 24 '16 at 5:01
  • $\begingroup$ Excuse me, but $f(x)=x^2$ should be concave function in $\mathbb R$? $\endgroup$ – Shine Mic Apr 24 '16 at 5:17
  • $\begingroup$ For any $x,y\in\mathbb R$, and $t\in[0,1]$, you have $$ (tx+(1-t)y)^2\leq tx^2+(1-t)y^2.$$ That makes it convex. $\endgroup$ – Martin Argerami Apr 24 '16 at 5:31

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