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A very hard prove

  1. Nov 21, 2007 #1
    hi,
    there is a challenging problem that no one could answer it,so here it is if any one wants to try:
    if x,z,y are all positive real numbers prove that:

    (1+x^2)(1+y^2)(1+z^2)/(xyz)>=8

    so anyone can help please? thank you.
     
  2. jcsd
  3. Nov 21, 2007 #2

    arildno

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    Hmm...you could try to start out with:
    [tex](1-x)^{2}(1-y)^{2}(1-z)^{2}\geq{0}[/tex]
    This can be rearranged as:
    [tex]((1+x^{2})-2x)((1+y^{2})-2y)((1+z^{2})-2z)\geq{0}[/tex]
    Perhaps this might yield something
     
  4. Nov 21, 2007 #3
    Hello,

    I am not a native english speaker, so be warned :-).
    This problem is simple really.
    For any real number we have
    (1-x)^2>=0, so 1+x^2>=2x;
    (1-y)^2>=0, so 1+y^2>=2y;
    (1-z)^2>=0, so 1+z^2>=2z.

    Lets plug these inequalities into the left part of inequality we want to prove:
    (1+x^2)(1+y^2)(1+z^2)/xyz >= (2x * 2y * 2z)/xyz=2*2*2=8.
    So, we have (1+x^2)(1+y^2)(1+z^2)/(xyz)>=8.
    Proved :-)
    Hope, this helps.
     
  5. Nov 21, 2007 #4

    arildno

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    Dear, oh dear, you're right. How simple..

    Here's another one:
    We may rewrite the the expression as:
    [tex]f(x,y,z)=g(x)*g(y)*g(z), g(x)=x+\frac{1}{x}[/tex]
    Thus, the minimum value will occur at:
    [tex]g'(x)=g'(y)=g'(z)=0\to{x}=y=z=1[/tex]
     
    Last edited: Nov 21, 2007
  6. Nov 21, 2007 #5

    arildno

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    Didn't the paniure give that proof already, Kummer?
     
  7. Nov 21, 2007 #6
    Yes he did. I should be more careful lest someone accuse me of stealing answers. I deleted my post otherwise it does not look good.
     
  8. Nov 21, 2007 #7

    arildno

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    Hmm..more likely, it was the not-bothering-reading-all-the-posts syndrome you were the victim of.

    I believe you are forgiven. :smile:
     
  9. Nov 21, 2007 #8
    it's simple, just assume that each one is 1 or more so it will work for all of the integers from 1 and above
     
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