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Math Help - Inequalities

  1. #1
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    AMGM Inequality help?

    Alright, here goes.

    Prove that
    abc > (a+b-c)(a-b+c)(-a+b+c)

    What I've done:

    By AM>GM property,
    (a+b+c)/3 > (abc)^(1/3)

    By AM>GM property, I also found that
    (a+b+c)/3 > ((a+b-c)(a-b+c)(-a+b+c))^(1/3)

    From here onwards I'm stuck, since it doesn't really make sense now.
    Need help here.

    Thanks in advance.
    Last edited by LegendWayne; April 11th 2009 at 02:20 AM.
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let x=a+b-c, \ y=a-b+c, \ z=-a+b+c

    Then a=\frac{x+y}{2}, \ b=\frac{x+z}{2}, \ c=\frac{y+z}{2}

    and the inequality becomes (x+y)(x+z)(y+z)\geq 8xyz

    Now apply three times AM\geq GM.
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  3. #3
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    Quote Originally Posted by red_dog View Post
    Now apply three times AM\geq GM.
    (x+y)(x+z)(y+z)\geq 8xyz

    I've pondered for a while.
    But ultimately I get this:

    Applying AM\geq GM,


    \frac{(2x+2y+2z)^3}{27} \geq (x+y)(x+z)(y+z) \geq 8xyz

    Sorry if I am abit slow on understanding.
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  4. #4
    MHF Contributor red_dog's Avatar
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    x+y\geq 2\sqrt{xy}

    x+z\geq 2\sqrt{xz}

    y+z\geq 2\sqrt{yz}

    Then, (x+y)(x+z)(y+z)\geq 8\sqrt{x^2y^2z^2}=8xyz
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  5. #5
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    My deepest gratitude for offering your solution.
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