# Suppose $f(x)$ is a nonnegative convex function in $[0,1]$, prove an inequality.

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!

• Where $f^2(x)\equiv f(f(x))$ or $[f(x)]^2$? – Kenny Lau Apr 24 '16 at 2:36
• @KennyLau The second one, most likely. – user228113 Apr 24 '16 at 2:36
• @KennyLau $\big(f(x)\big)^2\triangleq f^2(x)$ – Shine Mic Apr 24 '16 at 2:37
• Not sure if it will work, but you're in a probability space and you have a convex function, so try Jensen's inequality. – Rick Sanchez Apr 24 '16 at 4:03

Let $f(x)=x^2$. Then $$\int_0^1 f^2(x)\,dx=\int_0^1x^4\,dx=0.2,$$ while $$\frac43\,\left(\int_0^1 f(x)\,dx\right)^2=\frac43\,\left(\int_0^1x^2\,dx\right)^2=\frac4{27}=0.\overline{148}.$$
• Excuse me, but $f(x)=x^2$ should be concave function in $\mathbb R$? – Shine Mic Apr 24 '16 at 5:17
• 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. – Martin Argerami Apr 24 '16 at 5:31