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Thread: Uniqueness of Minimum

  1. #1
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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

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

    Then

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

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


    It is trivial to prove that the minimum occurs when $\displaystyle b=a^{p-1}$ if we consider the function (with b fixed) $\displaystyle g(a)=\frac{1}{p}a^{p}+\frac{1}{q}b^{q}-ab$. 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
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    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

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

    Then

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

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


    It is trivial to prove that the minimum occurs when $\displaystyle b=a^{p-1}$ if we consider the function (with b fixed) $\displaystyle g(a)=\frac{1}{p}a^{p}+\frac{1}{q}b^{q}-ab$. 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 $\displaystyle \ln (t)$ is strictly concave on $\displaystyle D=(0,\infty )$ (its second derivative is negative in this interval), this means that for any $\displaystyle x,y \in D$ and $\displaystyle s,t\in (0,1)$ with $\displaystyle x\neq y$ and $\displaystyle s+t=1$ we have $\displaystyle s\ln (x) +t\ln (y) < \ln (sx+ty)$. Taking $\displaystyle x=a^p,\ y=b^q,\ s=\frac{1}{p}$ and $\displaystyle t=\frac{1}{q}$ gives the inequality right away, and with the strict concavity we get that the equality is achieved iff $\displaystyle x=y$.
    Last edited by Jose27; Jul 21st 2010 at 04:04 PM.
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