Geometric Inequality: Prove √(2x)+√(2y)+√(2z)≤√(x+y)+√(y+z)+√(x+z)

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Homework Help Overview

The discussion revolves around proving a geometric inequality involving square roots of sums of variables representing side lengths of a triangle. The original poster attempts to show that the expression involving square roots of sums is less than or equal to another expression involving square roots of individual variables.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the Ravi transformation to rewrite the inequality. There are attempts to apply known inequalities like AM-GM and the rearrangement inequality, but these have not yielded success. Some participants suggest squaring the expressions as a potential approach.

Discussion Status

The discussion is ongoing, with participants exploring various methods to tackle the problem. There is an acknowledgment of the challenges faced in applying certain inequalities, and some guidance has been offered regarding the squaring of expressions and alternative approaches.

Contextual Notes

Participants are working under the constraints of proving the inequality within the context of triangle side lengths, which may influence the assumptions and methods considered.

dengulakungen
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Homework Statement


Let a,b and c be lengths of sides in a triangle, show that
√(a+b-c)+√(a-b+c)+√(-a+b+c)≤√a+√b+√c

The Attempt at a Solution


With Ravi-transformation the expressions can be written as

√(2x)+√(2y)+√(2z)≤√(x+y)+√(y+z)+√(x+z).

Im stuck with this inequality. Can´t find a way to use any known inequalities such as AM-GM or the rearrangement inequality.
 
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Yes I have tried to square the experisions, but without success. I will try it again.
 
dengulakungen said:
Yes I have tried to square the experisions, but without success. I will try it again.

Your alternative expression of the inequality is the way to go. Even simpler: verify that for any two numbers ##x,y>0## we have ##\sqrt{x+y} \geq \frac{1}{2} \sqrt{2x} + \frac{1}{2} \sqrt{2y}##. Again, have you tried squaring?
 
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