Background
Steele, in The Cauchy-Schwarz Master Class, states the general Arithmetic-Geometric Mean Inequality as follows (pg 23):
$a_{1}^{p_{1}}a_{2}^{p_{2}} \ldots a_{n}^{p_{n}} \leq p_{1}a_{1} + p_{2}a_{2} + \ldots + p_{n}a_{n}$ for non-negative real numbers $a_1,a_2, \ldots a_n$ and each sequence $p_1, p_2, \ldots, p_n$ of positive real numbers which sums to one.
Steele then discusses and presents Polya's "dream" proof which I've summarized below:
Let $A:=p_{1}a_{1} + p_{2}a_{2} + \ldots + p_{n}a_{n}$ and $G:=a_{1}^{p_{1}}a_{2}^{p_{2}} \ldots a_{n}^{p_{n}}$.
We have $1+x\leq e^{x}$ for all $x\in \mathbb{R}$. By change of variables we have: $x\leq e^{x-1}$ for all $x\in\mathbb{R}.$
Then $$\frac{a_{k}}{A}\leq e^{(\frac{a_{k}}{A})-1}$$ for $k \in \{1,2,\ldots, n\}$.
Thus we have $$\left(\frac{a_{1}}{A}\right)^{p_{1}}\left(\frac{a_{2}}{A}\right)^{p_{2}} \ldots \left(\frac{a_{n}}{A}\right)^{p_{n}} \leq \exp{\left(\left\{\sum_{k=1}^{n}p_{k}\frac{a_{k}}{A}\right\}-1\right)}=1.$$
After we move the multiples of $A$ to the righthand side and note that $p_{1}+p_{2}+\ldots + p_{n}=1$, the general Arithmetic-Geometric Mean Inequality is proven.
My Question
I can follow the proof and see that it is correct. But where do the $\frac{a_{k}}{A}$ terms come from?
Steele does discuss this but I don't understand his explanation.
My summary of his explanation is as follows:
If we apply $x\leq e^{x-1}$ when $x=a_{k}$ then we get $a_{k}\leq e^{a_{k}-1}$. By raising both sides to $p_{k}$ and multiplying, we get $G \leq \exp{(A-1)}$.
Similarly applying $x\leq e^{x-1}$ with $x=A$ we get $A\leq \exp{(A-1)}$.
So we have the double bound $$\max{\{A, G\}} \leq \exp{(A-1)}.$$ If either $A$ or $G$ is equal to $\exp{(A-1)}$ then we'll be able to turn this double bound into an inequality between $A$ and $G$, which we want. Steele says that if we think about trying to exploit this, we're led to the idea of normalization and so we introduce the $\frac{a_{k}}{A}$ terms.
This last part is what I don't follow. I'm not sure what Steele means by normalization here or how we're led to this from thinking about the case in which $A$ or $G$ is equal to $\exp{(A-1)}$. Can anyone help me understand what's going on here? Thanks!
Best Answer
In my opinion the idea to consider $\frac{a_k}{A}$ comes rather from "backwards working".