Proving Inequality: Expert Assistance for Your Inequality Proof

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    Inequality Proof
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SUMMARY

This discussion focuses on proving the inequality involving the function \( p(y) \) and its integrals, specifically the expression: \[ \int^\infty_0y^\frac{2(n-1)}{n}p(y)\,\mathrm{d}y\Big(\int^\infty_0y^\frac{1}{n}p(y)\,\mathrm{d}y\Big)^2>\int^\infty_0y^\frac{2}{n}p(y)\,\mathrm{d}y\Big(\int^\infty_0y^\frac{(n-1)}{n}p(y)\,\mathrm{d}y\Big)^2 \] for integer \( n \geq 2 \). The discussion highlights that when \( n = 2 \), both sides of the inequality are equal, indicating that the inequality does not hold in this case. The participants seek expert assistance in proving the inequality for other integer values of \( n \).

PREREQUISITES
  • Understanding of integral calculus, particularly improper integrals.
  • Familiarity with probability density functions and properties of \( p(y) \).
  • Knowledge of inequalities in mathematical analysis.
  • Basic understanding of mathematical notation and expressions involving limits.
NEXT STEPS
  • Research techniques for proving inequalities in mathematical analysis.
  • Study properties of integrals involving probability density functions.
  • Explore the implications of the Cauchy-Schwarz inequality in integral forms.
  • Learn about specific cases of inequalities for different integer values of \( n \).
USEFUL FOR

Mathematicians, students studying advanced calculus, and anyone involved in mathematical proofs or analysis of inequalities.

Aaron792
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Can anyone help me prove this inequality?
See upload~
 

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Prove inequality
\begin{equation}<br /> \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&gt;\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<br /> \nonumber<br /> \end{equation}

p(y)&gt;0 and \int^\infty_0p(y)\,\mathrm{d}y=1 for any y

n is an integer and n\ge2$
 
When n = 2, both of those inequalities are equal, thus the inequality isn't true for n = 2.
 

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