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Math Help - Inequality problem

  1. #1
    Newbie Mathroum's Avatar
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    Feb 2008
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    Question Inequality problem

    Could anyone help me solving this inequality?

    a,b,c are positive real numbers, proove that:

    [a/(b+c) + 1/2]*[b/(a+c) + 1/2]*[c/(a+b) + 1/2] >- +1

    Thank You


    I tried to do it with LaTEx:

    ( \frac{a}{b+c} + \frac{1}{2})*( \frac{b}{a+c} + \frac{1}{2})*( \frac{c}{a+b} + \frac{1}{2}) \geq 1

    OR


    [ \frac{a}{b+c} + \frac{1}{2}][ \frac{b}{a+c} + \frac{1}{2}][ \frac{c}{a+b} + \frac{1}{2}] \geq 1
    Last edited by Mathroum; February 11th 2008 at 03:25 PM.
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  2. #2
    Newbie Mathroum's Avatar
    Joined
    Feb 2008
    Posts
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    Got it!

    Hey I succeeded in prooving the inequality. The solution :

    \frac{(2a+b+c)(2b+a+c)(2c+a+b)}{8(b+c)(a+c)(a+b)} \geq 1

    when we multiply the whole inequality by 8(b+c)(a+c)(a+b) and get rid of brackets, we get :

    2a^3+2b^3+2c^3-a^2b-ab^2-b^2c-a^2c-ac^2-bc^2 \geq 0

    a^3+b^3-a^2b-ab^2 = (a+b)^3-4a^2b-4ab^2
    a^3+c^3-a^2c-ac^2 = (a+c)^3-4a^2c-4ac^2
    b^3+c^3-b^2c-bc^2 = (b+c)^3-4b^2c-4bc^2

    then :

    (a+b)^3-4ab(a+b) = (a+b)[(a+b)^2 - 4ab] = (a+b)(a-b)^2
    ... same for (a+c) and (b+c)....

    Finally :  (a+b) \geq 0 , (a-b)^2 \geq 0 \Rightarrow (a+b)(a-b)^2 \geq 0
    Last edited by Mathroum; February 12th 2008 at 10:41 AM.
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