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**AdvoTV** 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.