1. ## 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.

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 $x=y$.