A small problem

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  • #1
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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.
 

Answers and Replies

  • #2
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What are x, y, z? Real numbers? Positive reals?
 
  • #3
arildno
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Well, let's see:

Rewrite your equation as:
[tex](x,y,z)\cdot(1+y,1+z,1+x)=6\to||(x,y,z)||||(1+y,1+z,1+x)||\cos\phi=6[/tex]
whereby is implied:
[tex](x^{2}+y^{2}+z^{2})\sqrt{1+\frac{2x+2y+2z+3}{x^{2}+y^{2}+z^{2}}}\geq{6}[/tex]
Maybe this is usable?
 
  • #5
arildno
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That was a much better idea!!
 
  • #6
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 [Broken]
 
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  • #7
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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.
 
  • #8
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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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