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Math Help - Proofing

  1. #1
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    Proofing

    If a, b and c are positive real numbers, prove that:
    bc(b+c)+ca(c+a)+ab(a+b)≥6abc
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  2. #2
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    Quote Originally Posted by cedricc View Post
    If a, b and c are positive real numbers, prove that:
    bc(b+c)+ca(c+a)+ab(a+b)≥6abc
    Minimum positive real numbers for a, b and c will be 1.
    So the expression bc(b+c)+ca(c+a)+ab(a+b) becomes......?
    For any other values for a, b and c more one, the given expression will be......?
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  3. #3
    Super Member Bacterius's Avatar
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    Quote Originally Posted by sa-ri-ga-ma
    Minimum positive real numbers for a, b and c will be 1.
    What if a = 0 + \epsilon ?
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  4. #4
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    Quote Originally Posted by sa-ri-ga-ma View Post
    Minimum positive real numbers for a, b and c will be 1.
    So the expression bc(b+c)+ca(c+a)+ab(a+b) becomes......?
    For any other values for a, b and c more one, the given expression will be......?
    but how do i reason this??
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  5. #5
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    OK.
    Now expand the given expression, and divide it by abc. You get
    [b^2c + bc^2 + c^2a +ca^2 + a^2b + ab^2]/abc
    = b/a + c/a + c/b + a/b + a/c + b/c
    Now consider
    b/a + a/b = (b^2 + a^2)/ab = [(a+b)^2- 2ab]/ab = [(a+b)^2/ab] -2 > 0
    because [(a+b)^2/ab] is always greater than 2.
    Similarly you can proceed for other terms in the expressions.
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