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Math Help - Inequality 3

  1. #1
    Junior Member
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    Inequality 3

    If a,b,c>0 and a+b+c=3 prove that

     \frac{1}{a^{2}-a+3}+\frac{1}{b^{2}-b+3}+\frac{1}{c^{2}-c+3}\leq 1
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  2. #2
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    Hi,

    firstly, we can notice that f(x)=\frac{1}{x^2-x+3} is concave on [0,\frac{3}{2}] (by 2 derivations). So, if a,b, and c are in [0,\frac{3}{2}], we can apply the inegality of convexity, and the result is obvious.

    Now, if at least one of the 3 numbers is > \frac{3}{2}. Then, the 2 others are < \frac{3}{2}. Let's say that c> \frac{3}{2} and a< \frac{3}{2} and b< \frac{3}{2}, for example.
    Then f(c) \leq f(\frac{3}{2}) and f(a) \leq f(\frac{1}{2}) and f(b) \leq f(\frac{1}{2})(because f has a maxima in x=\frac{1}{2} and is decreasing on  [\frac{1}{2},3].

    finally, f(a)+f(b)+f(c) \leq 2f(\frac{1}{2})+f(\frac{3}{2})<1


    is there a mistake ?
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    I love your solution!
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