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Math Help - Uniqueness of Minimum

  1. #1
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    Dec 2009
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    Uniqueness of Minimum

    I'm having a small problem with the following lemma:

    Let a and b be any positive numbers, and let p and q be any positive numbers (necessarily greater than 1) satisfying

    \frac{1}{p}+\frac{1}{q}=1.

    Then

    \frac{1}{p}a^{p}+\frac{1}{q}b^{q}\geq{ab}

    with equality if and only if a^p=b^q.


    It is trivial to prove that the minimum occurs when b=a^{p-1} if we consider the function (with b fixed) g(a)=\frac{1}{p}a^{p}+\frac{1}{q}b^{q}-ab. The value of the function at the minimum is zero, and thus, this proves what comes after the statement after "then".

    The problem I am currently facing is how to reason/prove (hopefully in a rigorous fashion) that the minimum is unique.

    Any help is much appreciated. Thank you.
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  2. #2
    Super Member
    Joined
    Apr 2009
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    México
    Posts
    721
    Quote Originally Posted by AdvoTV View Post
    I'm having a small problem with the following lemma:

    Let a and b be any positive numbers, and let p and q be any positive numbers (necessarily greater than 1) satisfying

    \frac{1}{p}+\frac{1}{q}=1.

    Then

    \frac{1}{p}a^{p}+\frac{1}{q}b^{q}\geq{ab}

    with equality if and only if a^p=b^q.


    It is trivial to prove that the minimum occurs when b=a^{p-1} if we consider the function (with b fixed) g(a)=\frac{1}{p}a^{p}+\frac{1}{q}b^{q}-ab. The value of the function at the minimum is zero, and thus, this proves what comes after the statement after "then".

    The problem I am currently facing is how to reason/prove (hopefully in a rigorous fashion) that the minimum is unique.

    Any help is much appreciated. Thank you.
    The function \ln (t) is strictly concave on D=(0,\infty ) (its second derivative is negative in this interval), this means that for any x,y \in D and s,t\in (0,1) with x\neq y and s+t=1 we have s\ln (x) +t\ln (y) < \ln (sx+ty). Taking x=a^p,\ y=b^q,\ s=\frac{1}{p} and t=\frac{1}{q} gives the inequality right away, and with the strict concavity we get that the equality is achieved iff x=y.
    Last edited by Jose27; July 21st 2010 at 04:04 PM.
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