Show that the inequality is true | Geometric Mean

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Homework Help Overview

The discussion revolves around proving an inequality involving strictly positive numbers and their geometric mean. The inequality states that \((1+R_{G})^{n} \leq V\), where \(R_{G}\) is the geometric mean of the numbers and \(V\) is a product involving the numbers increased by one.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various methods to approach the problem, including taking the logarithm of both sides and expanding terms. There is mention of using the arithmetic mean-geometric mean inequality and a suggestion to relate the variables to exponential functions.

Discussion Status

The discussion is active with multiple approaches being explored. Some participants are considering different mathematical inequalities, while others are looking for additional insights or alternative methods. No consensus has been reached yet.

Contextual Notes

Participants note hints from the professor regarding the use of logarithmic functions and the relationship between the variables and exponential forms, indicating potential constraints or specific directions for the problem-solving process.

michonamona
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Homework Statement



Let r_{1}, r_{2}, ... , r_{n} be strictly positive numbers. Show that the inequality

(1+R_{G})^{n} \leq V

is true. Where R_{G} = (r_{1}r_{2}...r_{n})^{1/n} and V= \Pi_{k=1}^{n} (1+r_{k})

Homework Equations





The Attempt at a Solution



I've tried taking the log of both sides, as well as expanding out the term. Any insight?

Thanks,
M
 
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Use the arithmetic mean-geometric mean inequality... several times.
 
Any other insights?

The prof hinted that we should use log(1+e^x) and associate r with e^x.
 
That's an entirely different way to approach it. The approach I was thinking of uses the fact that the terms of the right side are the elementary symmetric functions of the r_j.
 

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