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!

#Algebra #Inequalities #AM-GMInequality #Polya #CauchySchwarz

Note by Ryan Tamburrino
6 years, 4 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

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, 4 months ago

Log in to reply

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

Jake Lai - 6 years, 4 months ago

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

Calvin Lin Staff - 6 years, 4 months ago

Log in to reply

Sure, I'd love to!

Ryan Tamburrino - 6 years, 4 months ago
×

Problem Loading...

Note Loading...

Set Loading...