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Math Help - Holder's Inequality

  1. #1
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    Holder's Inequality

    I need to prove that the following inequality is true only if t=0, t=1 or a=b>0:
    a^{t}b^{1-t}\leq ta+(1-t)b
    I have no idea how to proceed.
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  2. #2
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    The inequality is true for each 1<t<1 also! it is just the main ingredient for proving Holder inequality. Since log is incresing, the inequality is equivalent to
    log(a^tb^{1-t})\leq log(ta+(1-t)b)

    Now apply basic properties of the logarithm to the left hand side and use that log is concave to conclude.
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  3. #3
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    Im not sure how to apply the concept of concavity to this problem...
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by dannyboycurtis View Post
    Im not sure how to apply the concept of concavity to this problem...
    The concavity of log means that given two points a and b and the line \ell(t) going through \log(a) and \log(b), for any point c\in[a,b], \log(c)\geq\ell(c).

    Represent the point c as c=ta+(1-t)b for some t\in[0,1].
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