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Math Help - (1+x)...

  1. #1
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    Exclamation (1+x)...

    xyz=1

    (1+x)(1+y)(1+z) - 2 >= 2*(((x/y)^(1/3)) * ((y/z)^(1/3)) * ((z/x)^(1/3)) )

    How to proof it??
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ferrumes View Post
    xyz=1

    (1+x)(1+y)(1+z) - 2 >= 2*(((x/y)^(1/3)) * ((y/z)^(1/3)) * ((z/x)^(1/3)) )

    How to proof it??
    What are the restrictions? must (x,y,z)\in\mathbb{Z}^{+}?
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  3. #3
    GAMMA Mathematics
    colby2152's Avatar
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    Quote Originally Posted by ferrumes View Post
    xyz=1

    (1+x)(1+y)(1+z) - 2 >= 2*(((x/y)^(1/3)) * ((y/z)^(1/3)) * ((z/x)^(1/3)) )

    How to proof it??
    Note that the 2nd side simplifies to 2, so all you have left is:

    (x+1)(y+1)(y+z) \ge 4

    xyz-1
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  4. #4
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    Quote Originally Posted by Mathstud28 View Post
    What are the restrictions? must (x,y,z)\in\mathbb{Z}^{+}?
    yes
    [tex](x,y,z)\in\mathbb{Z}^{+}
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