# Talk:Inequality of arithmetic and geometric means

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Field:  Probability and statistics

## Untitled

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

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)

## 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 ${\displaystyle \ x_{a}x_{b}}$ 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 ${\displaystyle \sum _{i=1}^{n}x_{i}=n}$?

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 ${\displaystyle x_{a}x_{b}\leq 1}$ for all ${\displaystyle a,b}$ then ${\displaystyle x_{1}x_{2}\ldots x_{n}\leq 1}$.
2. Hmm, I have to think a bit about this one. It's not obvious to me; it needs some hypothesis on ${\displaystyle x_{a}}$ and ${\displaystyle x_{b}}$. 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, ${\displaystyle y_{1},y_{2},\ldots }$, where ${\displaystyle y_{i}=px_{i}}$. I'll make this change.
Oli Filth 10:22, 8 July 2007 (UTC)
${\displaystyle \ {\sqrt[{n}]{x_{1}\cdot x_{2}\cdots x_{n}}}\leq {\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}}$
to
${\displaystyle \ x_{1}\cdot x_{2}\cdots x_{n}\leq 1}$
in the general case (where ${\displaystyle \ \sum x_{i}\neq n}$)? Oli Filth 10:30, 8 July 2007 (UTC)

### Details for points 2 and 3

To point 2: First observe that

${\displaystyle {\frac {x_{a}x_{b}}{1+\alpha -\beta }}\leq 1}$

since

${\displaystyle 0.

In the proof we found already

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

By the induction hypothesis one gets

${\displaystyle x_{1}\cdots x_{a-1}x_{a+1}\cdots x_{b-1}x_{b+1}\cdots x_{k}(1+\alpha -\beta )\leq 1}$.

Multiplying by

${\displaystyle {\frac {x_{a}x_{b}}{1+\alpha -\beta }}\leq 1}$

gives

${\displaystyle x_{1}\cdots x_{a-1}x_{a}x_{a+1}\cdots x_{b-1}x_{b}x_{b+1}\cdots x_{k}\leq 1}$

which completes the proof of the lemma.

To point 3: Let ${\displaystyle u_{i}}$ be arbitrary positive values. Define

${\displaystyle a={\frac {\sum _{i=1}^{n}u_{i}}{n}}}$

and

${\displaystyle x_{i}={\frac {u_{i}}{a}}}$

Then

${\displaystyle \sum _{i=1}^{n}x_{i}=n}$

By the proved lemma one has

${\displaystyle 1\geq \prod _{i=1}^{n}x_{i}={\frac {\prod _{i=1}^{n}u_{i}}{a^{n}}}}$,

which is equivalent to

${\displaystyle a\geq {\sqrt[{n}]{\prod _{i=1}^{n}u_{i}}}}$. --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)

## 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?
${\displaystyle \prod _{i=1}^{n}u_{i}=1}$ implies ${\displaystyle \sum _{i=1}^{n}u_{i}\geq n}$
The proof can be found in de:Ungleichung vom arithmetischen und geometrischen Mittel and is essentially the same as the proof for
${\displaystyle \ x_{1}+\cdots +x_{n}=n\mu .\,}$ implies ${\displaystyle \ x_{1}\cdot x_{2}\cdots x_{n}\leq \mu ^{n}\,}$
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 (talkcontribs)

## feasibility of the geometric interpretation

Regarding the Geometric Interpretation: In the 2D case, 4 ${\displaystyle a\times b}$ rectangles always fit inside a ${\displaystyle (a+b)\times (a+b)}$ square. As I recall, this result has been extended to the 3d, 4d, and 5d cases. Does anyone know if it was extended beyond 5d? I think that a reference of the known results is in order, but I can't recall the papers I've read about it. Any help? mousomer (talk) —Preceding undated comment added 11:45, 7 January 2010 (UTC).

## Problem with formula in the Geometric interpretation section

In the section "Geometric interpretation" second paragraph. Shouldn't the first part/term of the formula for "total length of edges connected to a vertex on an n-dimensional cube" be :${\displaystyle 2n\,}$, ${\displaystyle n^{2}\,}$ or ${\displaystyle n^{2}\,}$ instead of just ${\displaystyle n\,}$. It is clear that this formula does not add up to the perimeter calculated for the square in the preceding paragraph. GreenManzanilla (talk) 16:22, 26 August 2010 (UTC)

The total number of edges is n2n−1. The was that is arrived at is this: n edges meet at each vertex and there are 2n vertices. This might suggest n2n, but that counts every edge twice, since every edge meets two vertices. Therefore, divide it by 2. Michael Hardy (talk) 17:45, 26 August 2010 (UTC)
....now how many edges of length x1 are there? To be continued.... Michael Hardy (talk) 17:47, 26 August 2010 (UTC)
Answer: There are just as many edges of length x1 as there are edges of any other length. Since there are n lengths and n2n−1 edges, there must be 2n−1 edges of each length. Michael Hardy (talk) 20:23, 26 August 2010 (UTC)
...and so the total edge length is 2n−1(x1 + ... + xn). Michael Hardy (talk) 20:25, 26 August 2010 (UTC)

....and now I've cleaned up that section. Michael Hardy (talk) 15:40, 27 August 2010 (UTC)

## Article name

When on earth is this ever called the "inequality of arithmetic and geometric means". Everyone calls it the AM–GM inequality. I could understand calling it the "arithmetic mean–geometric mean inequality" in full for clarity instead, but surely we should use one of these variants, rather than a made up Wikipedia-only name? Quietbritishjim (talk) 12:33, 27 June 2013 (UTC)

## Proof in lead section for n=2

Perhaps the proof currently in the lead section gives insight into the more general case, but it's so simple to prove it looks odd taking this slightly indirect route. Maybe it would be better to use the direct proof instead?

Current proof: This case can be seen from the elementary difference of squares formula: if ${\displaystyle a\geq b\geq 0,}$ then setting[note 1] ${\displaystyle x=(a+b)/2}$ and ${\displaystyle y=(a-b)/2,}$ so ${\displaystyle a=x+y}$ and ${\displaystyle b=x-y,}$ yields:
{\displaystyle {\begin{aligned}(a+b)/2&={\bigl (}(x+y)+(x-y){\bigr )}/2=x\\{\sqrt {ab}}&={\sqrt {(x+y)(x-y)}}={\sqrt {x^{2}-y^{2}}}\leq x\qquad {\text{equality if and only if }}y=0,{\text{ equivalently }}a=b.\end{aligned}}}
Direct proof: This can be seen from the fact that the square of a real number is always positive, so
${\displaystyle 0\leq (a-b)^{2}=a^{2}-2ab+b^{2}=a^{2}+2ab+b^{2}-4ab=(a+b)^{2}-4ab.}$
In other words ${\displaystyle (a+b)^{2}\geq 4ab}$, with equality precisely when ${\displaystyle a-b=0}$ i.e.${\displaystyle a=b}$. We get the result by square rooting both sides and using the fact that the square root function is strictly increasing.

Of course the two proofs are exactly the same, and the first one still uses the fact that the square root function is strictly increasing but just doesn't mention it. But I think the second wording is easier to a novice. Quietbritishjim (talk) 12:55, 27 June 2013 (UTC)

Many thanks to User:Schmock for making this update! Quietbritishjim (talk) 16:59, 28 June 2013 (UTC)

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