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Math Help - Prove (sqr(xy)+sqr(yz)+sqr(xz))^2 >= 3sqr(3)xyz with x + y + z = xyz

  1. #1
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    Post Prove (sqr(xy)+sqr(yz)+sqr(xz))^2 >= 3sqr(3)xyz with x + y + z = xyz

    Hi all,

    Please help to prove that: (sqr(xy)+sqr(yz)+sqr(xz))^2 >= 3sqr(3)xyz with: x, y, z are positive numbers and x + y + z = xyz.

    Thanks.
    Last edited by mr fantastic; March 21st 2011 at 06:30 PM. Reason: Re-titled.
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  2. #2
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    Indication:

    x, y, z > 0 \Rightarrow  (\exists )a, b, c \in \left ( -\frac{\pi }{2}, \frac{\pi }{2}\right ) such that  x=\tan a, y=\tan b, z=\tan c (tangent function is surjective)

    x+y+z=xyz\Rightarrow a+b+c=\pi (Prove that!)
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  3. #3
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    Cool

    Quote Originally Posted by veileen View Post
    Indication:

    x, y, z > 0 \Rightarrow (\exists )a, b, c \in \left ( -\frac{\pi }{2}, \frac{\pi }{2}\right ) such that  x=\tan a, y=\tan b, z=\tan c (tangent function is surjective)

    x+y+z=xyz\Rightarrow a+b+c=\pi (Prove that!)
    Thanks veileen very much.

    If you have another solution without using tan, please show me.
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  4. #4
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    Apologies if im missing something, but can you confirm this is the proposition:

    \left(\sqrt{xy} + \sqrt{yz} + \sqrt{xz}\right)^2 \geq 3(\sqrt{3}) xyz

    subject to:
    x,y,z >0
    xyz = x + y + z.


    There are plenty of counter examples, eg:
    x=500
    y = 1
    z = 501/499

    (xyz = x+ y + z = 502.004008 as required.)

    LHS = 2094.722
    RHS = 2608.489
    Last edited by SpringFan25; March 30th 2011 at 01:26 PM.
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  5. #5
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    Quote Originally Posted by SpringFan25 View Post
    There are plenty of counter examples,
    Agree...
    x=1, y=3/2, z=5 :
    xyz = x+y+z = 15/2

    LHS = 38.432878...
    RHS = 38.971143...
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