Prove that (a+b)(b+c)(c+a) =/> 8abc

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Discussion Overview

The discussion centers around proving the inequality \((a+b)(b+c)(c+a) \geq 8abc\) for non-negative values of \(a\), \(b\), and \(c\). Participants explore various mathematical approaches and reasoning related to this inequality.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests ordering \(a\), \(b\), and \(c\) from largest to smallest and expanding the product to analyze the terms.
  • Another participant questions whether \(abc\) is greater than or equal to \(b^2c\), implying a potential flaw in the previous reasoning.
  • A suggestion is made to apply the AM-GM inequality to the terms involved in the expansion.
  • There is a clarification regarding the interpretation of the notation \(\geq\) as "equal or more".
  • One participant provides an expanded form of the inequality and proposes using the GM-AM inequality to establish a lower bound involving the geometric mean of the terms.

Areas of Agreement / Disagreement

Participants express differing views on the validity of certain steps in the proof and the application of inequalities, indicating that the discussion remains unresolved with multiple competing approaches.

Contextual Notes

Some assumptions about the non-negativity of \(a\), \(b\), and \(c\) are made, but the implications of negative values are briefly mentioned without resolution. The discussion includes various mathematical steps that are not fully resolved.

chinyew
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prove that (a+b)(b+c)(c+a) =/> 8abc
for all a,b,c =/> 0
any1 pls.. thx.
 
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Let's start by ordering them from largest to smallest: a \ge b \ge c.
Then you can open the brackets and get 8 terms, two of which are precisely equal to abc. Picking one at random, say, b2c, can you show that this is larger than abc?
 
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abc≥b2c, so the answer to your question is no.
 


You should use AM-GM inequality which
\sqrt{ab} </= (a+b)/2
 


Given a,b,c =/> 0 implies
a,b,c < 0 implies
(a+b) < a
(b+c) < b
(c+a) < c
...
 


He was meant to say that =/> is read "equal or more".

Compuchip's approach will be easiest.
 


When expanded you get

a^{2}b + a^{2}c + ab^{2} + ac^{2} + b^{2}c + bc^2 + abc + abc \geq 8abc

If you apply GM-AM inequality to the collection you get:

a^{2}b + a^{2}c + ab^{2} + ac^{2} + b^{2}c + bc^2 + abc + abc \geq 8\sqrt[8]{a^{8}b^{8}c^{8}}
 

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