# Help inequality proof

1. May 13, 2008

### Aaron792

Can anyone help me prove this inequality?

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2. May 14, 2008

### Aaron792

A complete description
Prove inequality
$\int^\infty_0y^\frac{2(n-1)}{n}p(y)\,\mathrm{d}y\Big(\int^\infty_0y^\frac{1}{n}p(y)\,\mathrm{d}y\Big)^2>\int^\infty_0y^\frac{2}{n}p(y)\,\mathrm{d}y\Big(\int^\infty_0y^\frac{(n-1)}{n}p(y)\,\mathrm{d}y\Big)^2 \nonumber$

$p(y)>0$ and $\int^\infty_0p(y)\,\mathrm{d}y=1$ for any y

n is an integer and $n\ge2$

3. May 14, 2008

### sennyk

When n = 2, both of those inequalities are equal, thus the inequality isn't true for n = 2.