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Thread: an inequality

  1. #1
    Super Member Rebesques's Avatar
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    an inequality

    Here's a little troublesome inequality: Prove that

    \alpha^\beta+\beta^\alpha> 1,  <br />
  \forall  \  0<\alpha,\beta<1.

    I would appreciate a shorter solution, but with less theory involved... Asking 2 much?
    Last edited by Rebesques; Jul 25th 2005 at 08:30 AM.
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  2. #2
    Super Member Rebesques's Avatar
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    :(

    Just noticed, this is the wrong place for the topic to be... Could one of the administrators please move this thread to the "inequalities" section?
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  3. #3
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    The inequalities forum is for elementary school math (for younger students). If you don't want it in the university level algebra forum, let me know which section suits it the best.
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  4. #4
    hpe
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    Quote Originally Posted by Rebesques
    Prove that
    \alpha^\beta+\beta^\alpha> 1,  <br />
  \forall  \  0<\alpha,\beta<1.
    Fix  0< \beta < 1 and consider the function f(\alpha) = \alpha^\beta + \beta^\alpha = g(\alpha) + h(\alpha).

    Then f(0) = 1, f(1) = 1+\beta, and both g(x) and h(x) are log-concave (that is, their logarithms are concave down). It is known that the sum of log-concave functions is also log-concave, so f is log-concave. As a consequence,
    \alpha^\beta + \beta^\alpha = f(\alpha) \ge f(0)^{1-\alpha}f(1)^\alpha = (1+\beta)^\alpha > 1.
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  5. #5
    Super Member Rebesques's Avatar
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    ?

    Ellegant, short, not heavily theoretic; Excellent

    Thanks!
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