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Math Help - inequalty

  1. #1
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    inequalty

    if a,b,c > 0 and a+b+c=1

    prove

    \frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}  {c^2+a^2}\geq\frac{1}{ 2}
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  2. #2
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    Re: inequalty

    Solution:
    Given a,b,c>0
    a+b+c=1
    this is possible when a=b=c=1/3
    a^3/(a+b)+b^3/(b+c)+c^3/(c+a)>=1/2
    then (1/3)/(1/9+1/9)+ (1/3)/(1/9+1/9)+(1/3)/(1/9+1/9)
    =1/2
    hence proved
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  3. #3
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    Re: inequalty

    Quote Originally Posted by brosnan123 View Post
    Solution:
    Given a,b,c>0
    a+b+c=1
    this is possible when a=b=c=1/3
    a^3/(a+b)+b^3/(b+c)+c^3/(c+a)>=1/2
    then (1/3)/(1/9+1/9)+ (1/3)/(1/9+1/9)+(1/3)/(1/9+1/9)
    =1/2
    hence proved
    This is not a proof, you only showed it was true for one set of values for a, b and c

    Edit: I didn't realise this question was from 2008.
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