Calling All Cauch Polyatatoes

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 aka_k with 1kn1\leq k\leq n and a sequence of reals pkp_k with 1kn1\leq k \leq n such that k=1npk=1\sum_{k=1}^{n} p_k = 1, that

a1p1a2p2a3p3...anpna1p1+a2p2+a3p3+...+anpna_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+xex1+x\leq e^x which can easily be seen graphically with equality occurring only at x=0x=0. If we make the change of variables xx1x\mapsto x-1, our initial observation becomes xex1x\leq e^{x-1}. Applying this to our sequence aka_k we get akeak1a_k\leq e^{a_k -1}

akpkeakpkpka_k^{p_k}\leq e^{a_kp_k - p_k} now, we can see that G=k=1nakpkexp(k=1nakpkk=1npk)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=k=1nakpkexp(k=1nakpk1)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=k=1nakpkexp(k=1nakpk1)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 GG. So we could say

A,Gexp(k=1nakpk1)A, G \leq \exp \left (\sum_{k=1}^{n} a_kp_k -1 \right ) So we have related AA and GG by inequality, but we have not separated them. Now we look closer at the case where AA and GG are equal to the expression on the right. The idea of "normalization" then comes to mind. Normalization was discussed earlier in the book, but this is precisely where I got lost.\color{limegreen}{\text{this is precisely where I got lost.}} What Steele does is the following:

Define a new sequence αk\alpha_k with 1kn1\leq k \leq n where we have αk=akA \alpha_k = \frac{a_k}{A} where we have A=a1p1+a2p2+a3p3+...+anpnA=a_1p_1+a_2p_2+a_3p_3+...+a_np_n

Applying our earlier bound for GG for our new variables αk\alpha_k, we get k=1nαkpkexp(k=1nαkpk1)\prod_{k=1}^{n} \alpha_k^{p_k} \leq \exp \left(\sum_{k=1}^{n} \alpha_kp_k-1\right) k=1n(akA)pkexp(k=1nakApk1)\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 k=1nakApk=1\sum_{k=1}^{n} \frac{a_k}{A}p_k = 1 and it follows that exp(k=1nakApk1)=1\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 αk\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!

Note by Ryan Tamburrino
6 years, 6 months ago

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Update: 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.

k=1nakApk=a1p1+a2p2+a3p3+...+anpnA=AA=1\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.

Ryan Tamburrino - 6 years, 6 months ago

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

Jake Lai - 6 years, 6 months ago

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

Calvin Lin Staff - 6 years, 6 months ago

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Sure, I'd love to!

Ryan Tamburrino - 6 years, 6 months ago

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