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Math Help - An algebric inequality

  1. #1
    Senior Member bkarpuz's Avatar
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    An algebric inequality

    For x,y,z>0, prove
    x^{5}+y^{5}+z^{5}\leq x^{5}\sqrt{\frac{x^{2}}{yz}}+y^{5}\sqrt{\frac{y^{2  }}{xz}}+z^{5}\sqrt{\frac{z^{2}}{xy}}.

    thanks.
    Last edited by bkarpuz; November 11th 2008 at 12:09 AM. Reason: \geq\to\leq
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  2. #2
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    Quote Originally Posted by bkarpuz View Post
    For x,y,z>0, prove
    x^{5}+y^{5}+z^{5}\geq x^{5}\sqrt{\frac{x^{2}}{yz}}+y^{5}\sqrt{\frac{y^{2  }}{xz}}+z^{5}\sqrt{\frac{z^{2}}{xy}}.

    thanks.
    the inequality in this form is not true for all positive reals x, y, z, because it fails if you put z = 1, x = y, and then let x go to infinity! are you sure it's not \leq instead of \geq ?
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  3. #3
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    the inequality in this form is not true for all positive reals x, y, z, because it fails if you put z = 1, x = y, and then let x go to infinity! are you sure it's not \leq instead of \geq ?
    I correct it now.
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    Quote Originally Posted by bkarpuz View Post
    For x,y,z>0, prove
    x^{5}+y^{5}+z^{5}\leq x^{5}\sqrt{\frac{x^{2}}{yz}}+y^{5}\sqrt{\frac{y^{2  }}{xz}}+z^{5}\sqrt{\frac{z^{2}}{xy}}.

    thanks.
    because of symmetry, we may assume that x \leq y \leq z. then obviously: \sqrt{\frac{x^{2}}{yz}} \leq \sqrt{\frac{y^{2}}{xz}} \leq \sqrt{\frac{z^{2}}{xy}}. hence by Chebyshev's inequality we'll have:

    x^{5}\sqrt{\frac{x^{2}}{yz}}+y^{5}\sqrt{\frac{y^{2  }}{xz}}+z^{5}\sqrt{\frac{z^{2}}{xy}} \geq \frac{1}{3}(x^5 + y^5 + z^5) \left(\sqrt{\frac{x^{2}}{yz}}+\sqrt{\frac{y^{2}}{x  z}}+\sqrt{\frac{z^{2}}{xy}} \right). but by AM-GM we have: \sqrt{\frac{x^{2}}{yz}}+\sqrt{\frac{y^{2}}{xz}}+\s  qrt{\frac{z^{2}}{xy}} \geq 3. \ \ \ \ \ \Box


    imagine how many inequalities you can create using only Chebyshev's inequality + AM-GM inequality! i guess i'm back to high school again!
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    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    because of symmetry, we may assume that x \leq y \leq z. then obviously: \sqrt{\frac{x^{2}}{yz}} \leq \sqrt{\frac{y^{2}}{xz}} \leq \sqrt{\frac{z^{2}}{xy}}. hence by Chebyshev's inequality we'll have:

    x^{5}\sqrt{\frac{x^{2}}{yz}}+y^{5}\sqrt{\frac{y^{2  }}{xz}}+z^{5}\sqrt{\frac{z^{2}}{xy}} \geq \frac{1}{3}(x^5 + y^5 + z^5) \left(\sqrt{\frac{x^{2}}{yz}}+\sqrt{\frac{y^{2}}{x  z}}+\sqrt{\frac{z^{2}}{xy}} \right). but by AM-GM we have: \sqrt{\frac{x^{2}}{yz}}+\sqrt{\frac{y^{2}}{xz}}+\s  qrt{\frac{z^{2}}{xy}} \geq 3. \ \ \ \ \ \Box


    imagine how many inequalities you can create using only Chebyshev's inequality + AM-GM inequality! i guess i'm back to high school again!
    I was thinking of a different solution.
    I know that this inequality can be shown by special inequalities, the first one dropped in my mind was Muirhead's inequality (again because of the symmetry), which you may find in Muirhead's inequality - Wikipedia, the free encyclopedia.
    If there is any other solutions, I would like to see.

    Thanks NonCommAlg.
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    Quote Originally Posted by bkarpuz View Post
    I was thinking of a different solution.
    I know that this inequality can be shown by special inequalities, the first one dropped in my mind was Muirhead's inequality (again because of the symmetry), which you may find in Muirhead's inequality - Wikipedia, the free encyclopedia.
    If there is any other solutions, I would like to see.

    Thanks NonCommAlg.
    when i was a high school kid, we were trained to avoid using Muirhead's inequality as much as we can! i never knew why? i guess i'm still under the infulence! lol

    anyway, yes, Muirhead's inequality will also solve the problem because if you multiply both sides of the inequality by xyz you'll get: \sum_{sym}x^6yz \leq \sum_{sym}x^7 \sqrt{y}\sqrt{z},

    which is true because 7, \frac{1}{2}, \frac{1}{2} majorizes 6, 1, 1.
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