Hello all! I was recently reading a proof of the AM-GM inequality by George Polya in my book titled "The Cauchy-Schwarz Master Class" by professor J. Michael Steele. I think it's an extraordinary, beautiful proof, but there's one step that I don't quite understand. I'll post a relatively quick proof, pointing out my area of confusion when I get there.

So, we want to prove that for a sequence of reals $a_k$ with $1\leq k\leq n$ and a sequence of reals $p_k$ with $1\leq k \leq n$ such that $\sum_{k=1}^{n} p_k = 1$, that

$a_1^{p_1}a_2^{p_2}a_3^{p_3}...a_n^{p_n}\leq a_1p_1+a_2p_2+a_3p_3+...+a_np_n$

Polya's proof begins with the observation that $1+x\leq e^x$ which can easily be seen graphically with equality occurring only at $x=0$. If we make the change of variables $x\mapsto x-1$, our initial observation becomes $x\leq e^{x-1}$. Applying this to our sequence $a_k$ we get $a_k\leq e^{a_k -1}$

$a_k^{p_k}\leq e^{a_kp_k - p_k}$ now, we can see that $G=\prod_{k=1}^{n} a_k^{p_k} \leq \exp \left(\sum_{k=1}^{n} a_kp_k - \sum_{k=1}^{n} p_k\right)$ $G=\prod_{k=1}^{n} a_k^{p_k} \leq \exp \left(\sum_{k=1}^{n} a_kp_k - 1\right)$ however, we can also see from one of our initial observation that $A=\sum_{k=1}^{n} a_kp_k \leq \exp \left(\sum_{k=1}^{n} a_kp_k -1\right)$ which is the same bound we just found for $G$. So we could say

$A, G \leq \exp \left (\sum_{k=1}^{n} a_kp_k -1 \right )$ So we have related $A$ and $G$ by inequality, but we have not separated them. Now we look closer at the case where $A$ and $G$ are equal to the expression on the right. The idea of "normalization" then comes to mind. Normalization was discussed earlier in the book, but $\color{LimeGreen}{\text{this is precisely where I got lost.}}$ What Steele does is the following:

Define a new sequence $\alpha_k$ with $1\leq k \leq n$ where we have $\alpha_k = \frac{a_k}{A}$ where we have $A=a_1p_1+a_2p_2+a_3p_3+...+a_np_n$

Applying our earlier bound for $G$ for our new variables $\alpha_k$, we get $\prod_{k=1}^{n} \alpha_k^{p_k} \leq \exp \left(\sum_{k=1}^{n} \alpha_kp_k-1\right)$ $\prod_{k=1}^{n} \left(\frac{a_k}{A}\right)^{p_k} \leq \exp \left(\sum_{k=1}^{n} \frac{a_k}{A}p_k-1\right)$ Now, Steele implies that $\sum_{k=1}^{n} \frac{a_k}{A}p_k = 1$ and it follows that $\exp \left(\sum_{k=1}^{n} \frac{a_k}{A}p_k-1\right) = 1$

Now from here, showing that the AM-GM inequality holds is easy. The only place that brings me confusion is the purpose of introducing and defining the sequence $\alpha_k$ in the way that Steele does. If someone could help me out with this, it would be greatly appreciated! Consider this a recommendation for "The Cauchy-Schwarz Master Class," it's a fantastic book and you'd all love it!

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestUpdate: a few minutes later, while proofreading the post for errors, I looked at what I was confused with and I have seen the light! I now understand why the sequence was introduced.

$\sum_{k=1}^{n} \frac{a_k}{A}p_k = \frac{a_1p_1+a_2p_2+a_3p_3+...+a_np_n}{A} = \frac{A}{A} = 1$

Seems so obvious now, but I guess that's how it always goes. I will leave the post and proof up because I think it's a great proof for the more general version of the AM-GM Inequality. Also, I feel like this is a prime example of how sometimes, one just needs to take a step back and look at problems from a fresh perspective.

Log in to reply

A bit of a confusing point for me too, but fantastic proof nonetheless.

Log in to reply

This is great! Can you add this to the AM-GM wiki? Thanks!

Log in to reply

Sure, I'd love to!

Log in to reply