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- Thread starter Aaron792
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- #2

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Prove inequality

[itex]\begin{equation}

\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

\end{equation}[/itex]

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

n is an integer and [itex]n\ge2$[/itex]

- #3

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

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