Can anyone help me prove this inequality? See upload~
May 13, 2008 #1 Aaron792 5 0 Can anyone help me prove this inequality? See upload~ Attachments 不等式.bmp 156.5 KB · Views: 443
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May 14, 2008 #2 Aaron792 5 0 A complete description 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$[/itex]
A complete description Prove inequality [itex]\begin{equation} \int^\infty_0y^\frac{2(n-1)}{n}p(y)\,\mathrm{d}y\Big(\int^\infty_0y^\frac{1}{n}p(y)\,\mathrm{d}y\Big)^2>\int^\infty_0y^\frac{2}{n}p(y)\,\mathrm{d}y\Big(\int^\infty_0y^\frac{(n-1)}{n}p(y)\,\mathrm{d}y\Big)^2 \nonumber \end{equation}[/itex] [itex]p(y)>0 [/itex] and [itex]\int^\infty_0p(y)\,\mathrm{d}y=1[/itex] for any y n is an integer and [itex]n\ge2$[/itex]
May 14, 2008 #3 sennyk 73 0 When n = 2, both of those inequalities are equal, thus the inequality isn't true for n = 2.