Talk:Inequality of arithmetic and geometric means

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umm, in the polya proof, the meaning of mu and rho are not given —Preceding unsigned comment added by 128.36.86.60 (talk • contribs)

But they are, in the second paragraph, where it says:
Let μ be the arithmetic mean, and let ρ be the geometric mean.
Michael Hardy 22:07, 11 December 2006 (UTC)

[edit] Proof by induction

Perhaps I'm just being slow today, but isn't the proof by induction section somewhat lacking?

  1. Why are we "done if we show that the product \ x_a x_b is less than one"?
  2. Where is the proof that this is "obviously true"?
  3. Even if this proof is correct, doesn't it only cover the case when \sum_{i=1}^n x_i = n?

Oli Filth 14:52, 1 July 2007 (UTC)

I numbered your questions so that my answers will correspond with them; hope you don't mind.
  1. If x_a x_b \le 1 for all a,b then x_1x_2\ldots x_n \le 1.
  2. Hmm, I have to think a bit about this one. It's not obvious to me; it needs some hypothesis on xa and xb. At first sight, this seems to be a potentially serious gap in the proof.
  3. No, that's what the business of dividing by p is about.
It's a bit terse, I know, but if point 2 is not resolved then the whole proof is wrong. I need another look (little time now). -- Jitse Niesen (talk) 08:30, 8 July 2007 (UTC)
Thanks for looking into this. Here are my followup remarks:
  1. I see. But in that case, why is it necessary to do the substitution stuff?
  2. ...
  3. Ah, I finally see what that division is trying to achieve. It would be much clearer if the proof was written in terms of, say, y_1, y_2, \ldots, where yi = pxi. I'll make this change.
Oli Filth 10:22, 8 July 2007 (UTC)
On second thought, I'm still not convinced about this! How do we get from:
\ \sqrt[n]{x_1 \cdot x_2 \cdots x_n} \leq \frac{x_1 + x_2 + \cdots + x_n}{n}
to
\ x_1 \cdot x_2 \cdots x_n \le 1
in the general case (where \ \sum x_i \neq n)? Oli Filth 10:30, 8 July 2007 (UTC)

[edit] Details for points 2 and 3

To point 2: First observe that

\frac{x_a x_b}{1+\alpha -\beta }\leq 1

since

0<x_a x_b = 1+\alpha -\beta -\alpha\beta\leq 1+\alpha -\beta .

In the proof we found already

\ x_1 + \cdots + x_{a-1} + x_{a+1} + \cdots + x_{b-1} + x_{b+1} + \cdots + x_k + (1+\alpha - \beta) = k

By the induction hypothesis one gets

x_1\cdots x_{a-1} x_{a+1} \cdots x_{b-1} x_{b+1} \cdots x_k (1+\alpha - \beta)\leq 1.

Multiplying by

\frac{x_a x_b}{1+\alpha -\beta }\leq 1

gives

x_1\cdots x_{a-1} x_a x_{a+1} \cdots x_{b-1} x_b x_{b+1} \cdots x_k \leq 1

which completes the proof of the lemma.

To point 3: Let ui be arbitrary positive values. Define

a=\frac{\sum_{i=1}^n u_i}{n}

and

x_i=\frac{u_i}{a}

Then

\sum_{i=1}^n x_i=n

By the proved lemma one has

1\geq \prod_{i=1}^n x_i=\frac{\prod_{i=1}^n u_i}{a^n},

which is equivalent to

a\geq \sqrt[n]{\prod_{i=1}^n u_i}. --NeoUrfahraner 11:01, 12 July 2007 (UTC)

I introduced the proof again in a corrected version - I hope you agree. --NeoUrfahraner 21:15, 16 July 2007 (UTC)

Looks fine to me. Thanks for the clarification. -- Jitse Niesen (talk) 04:15, 23 July 2007 (UTC)

[edit] proof by Newman

pf:

1st step: if AM-Gm holds for all nonnegative numbers such that its product is 1, then it holds for all nonnegative numbers.
2nd step: show that AM-GM holds for all nonnegative numbers such that its product it 1, by first showing that its sum is greater than ()
third step show
ok, i seriously need someone to tell me how do you type mathematics on internet

21:20, 26 August 2007 (UTC)21:20, 26 August 2007 (UTC)~~

You mean this?
\prod_{i=1}^n u_i =1 implies \sum_{i=1}^n u_i \geq n
The proof can be found in de:Ungleichung vom arithmetischen und geometrischen Mittel and is essentially the same as the proof for
\ x_1 + \cdots + x_n = n\mu.\, implies \ x_1 \cdot x_2 \cdots x_n \le \mu^n\,
that you find in this article. Have you any source that this proof is actually from Max Newman? --NeoUrfahraner 10:34, 5 September 2007 (UTC)


Yes, I think that's it. There are at least five different proofs of this inequality out there, at least I know five. I am not sure if this particular one can be attributed to whom. I am not even sure if we are talking about the same Newman! —Preceding unsigned comment added by 128.226.170.133 (talk • contribs)