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Math Help - prove <= 8/27

  1. #1
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    prove <= 8/27

    for a,b,c real positive numbers, show that

    ((a+b)(b+c)(c+a))/ ((a+b+c)^3) <= 8/27

    i'm thinking of cuberoot both sides..what do u guys think?
    or is there any theorem?
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  2. #2
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    hmm i'm not sure it's true... this is the route i went down:

    let d = max(a,b,c) and e = min(a,b,c)

    then

    {your LHS} <= (d+d)(d+d)(d+d)/(e+e+e)^3 = ((2d)^3)/((3e)^3)

    =8d^3/27e^3 = (8/27) * (d/e)^3

    But e <= d so the upper bound can be more than 8/27...

    You sure its true?

    Si
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  3. #3
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by nh149 View Post
    for a,b,c real positive numbers, show that

    ((a+b)(b+c)(c+a))/ ((a+b+c)^3) <= 8/27
    By AM–GM,

    2(a+b+c)

    =\ (a+b)+(b+c)+(c+a)

    \ge\ 3\sqrt[3]{(a+b)(b+c)(c+a)}

    The result follows by cubing both sides.
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