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A small problem

  1. Feb 17, 2008 #1
    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.
     
  2. jcsd
  3. Feb 17, 2008 #2
    What are x, y, z? Real numbers? Positive reals?
     
  4. Feb 17, 2008 #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. Feb 17, 2008 #4
  6. Feb 17, 2008 #5

    arildno

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    That was a much better idea!!
     
  7. Feb 17, 2008 #6
  8. Feb 17, 2008 #7
    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.
     
  9. Feb 18, 2008 #8
    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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