Proving x^2+y^2+z^2 is ≥3 Given x+y+z+xy+yz+zx=6

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Discussion Overview

The discussion revolves around proving the inequality x² + y² + z² ≥ 3 given the constraint x + y + z + xy + yz + zx = 6. Participants explore various mathematical approaches and techniques to tackle this problem, including the use of Lagrange multipliers and the Cauchy-Schwarz inequality.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant requests hints for proving the inequality under the given constraint.
  • Another participant questions the nature of the variables x, y, and z, asking if they are real numbers or positive reals.
  • A participant suggests rewriting the equation in a vector form and implies a relationship involving the cosine of an angle, proposing that this might be useful for the proof.
  • Another participant proposes using Lagrange multipliers to find the extremums of x² + y² + z² under the given constraint, suggesting that verifying the minimum could yield the value 3.
  • A later reply expresses appreciation for the Lagrange multiplier approach, indicating it is a better idea.
  • One participant mentions an alternative method related to Lagrange multipliers and shares a link for further reading.
  • A participant shares a personal connection to the source of the alternative method, noting their studies in physics and mathematics at a university in Denmark.
  • Another participant suggests that the problem can be solved using the Cauchy-Schwarz inequality, providing inequalities for each variable that could lead to the solution.

Areas of Agreement / Disagreement

Participants express various approaches to the problem, with no consensus on a single method or solution. Multiple competing views remain regarding the best technique to use for the proof.

Contextual Notes

Some assumptions about the nature of the variables (e.g., whether they are real or positive reals) remain unresolved. The discussion includes references to mathematical techniques that may require further exploration to fully understand their applicability.

pixel01
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I do not know if this is suitable here:

Given that : x+y+z+xy+yz+zx=6,
Prove that x^2+y^2+z^2 >=3

Any hints will be appreciated. Thanks.
 
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What are x, y, z? Real numbers? Positive reals?
 
Well, let's see:

Rewrite your equation as:
(x,y,z)\cdot(1+y,1+z,1+x)=6\to||(x,y,z)||||(1+y,1+z,1+x)||\cos\phi=6
whereby is implied:
(x^{2}+y^{2}+z^{2})\sqrt{1+\frac{2x+2y+2z+3}{x^{2}+y^{2}+z^{2}}}\geq{6}
Maybe this is usable?
 
That was a much better idea!
 
mrandersdk said:
maybe use lagrange multiplier to find the extremums of x^2+y^2+z^2 under the constraint x+y+z+xy+yz+zx=6, then verify that the minimum gives the value 3.

see:

http://en.wikipedia.org/wiki/Lagrange_multipliers

I'll have to read up on that method. ANother alternative:

http://www.imf.au.dk/da/matematiklaererdag/2005/filer/niels.pdf
 
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that is a funny coincidens John Creighto I'm studying physics and mathematics at the university of Aarhus denmark, where your note is from.Haven't stydied the note though.
 
Thanks for all of your contributions.
This one can be solved using Cauchy though.
We have: x^2 + 1 >= 2x; y^2+1>=2y and z^2+1>=2z.
Then it's easy for the rest.
 

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