Talk:AM–GM inequality

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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)

Proof by averaging the arithmetic mean

 15:30, 24 March 2022 (UTC)
This section can be improved. Specifically, the parts about: (a) the replacement strategy for xi and xj not converging in a finite number of steps, and (b) the demonstration that the replacement strategy does converge in the limit to alpha, while interesting are largely irrelevant since a different (simpler?) replacement strategy will achieve the result while avoiding the convergence issues. I mean the following.

   If not all numbers are equal, then there exist i, j such that xi < alpha < xj. Then replacing xi by alpha and xj by xj + xi - alpha will leave the arithmetic mean on left-hand side unchanged, but will increase the geometric mean on the right-hand side because alpha (xj + xi - alpha) - xi xj = (alpha - xi)(xj - alpha) > 0.

 Clearly, after at most (n-1) such replacement steps, all the xi's will be replaced by alpha while each step maintains the arithmetic mean at alpha and increases the geometric mean by a positive amount. After the last step, the geometric mean equals arithmetic mean, thus proving the inequality.

 Mutatis mutandis, it can be noted that the replacement strategy can of course be done from the right-hand side as well. Specifically,

  If not all numbers are equal, then there exist i, j such that xi < beta < xj where beta is the geometric mean. Then replacing xi by beta and xj by (xi xj / beta) leaves the geometric mean on the right-hand side unchanged at beta. However, it decreases the arithmetic mean because xi + xj - beta - (xi xj / beta) = (beta - xi)(xj - beta)/beta > 0. 

--Litmus58 (talk) 15:30, 24 March 2022 (UTC)

I have now edited the article to make the changes mentioned above. Please note that I have also altered the section title since it better reflects the changes made.

Litmus58 (talk) 19:24, 28 March 2022 (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 $\ x_{a}x_{b}$ 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 $x_{a}x_{b}\leq 1$ for all $a,b$ then $x_{1}x_{2}\ldots x_{n}\leq 1$.
2. Hmm, I have to think a bit about this one. It's not obvious to me; it needs some hypothesis on $x_{a}$ and $x_{b}$. 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, $y_{1},y_{2},\ldots$, where $y_{i}=px_{i}$. 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

$\ x_{1}\cdot x_{2}\cdots x_{n}\leq 1$

in the general case (where $\ \sum x_{i}\neq n$)? Oli Filth 10:30, 8 July 2007 (UTC)

Details for points 2 and 3

To point 2: First observe that

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

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 $u_{i}$ 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)

Final Sentence of Proof by Induction Subsection 2

The final sentence of this subsection had "${=1}$" for the values $a_{i}$ even though $a_{i}$ should be able to take on any positive real value. I removed the "${=1}$".JasonHeine (talk) 23:12, 19 January 2020 (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?

$\prod _{i=1}^{n}u_{i}=1$ 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 $\ x_{1}\cdot x_{2}\cdots x_{n}\leq \mu ^{n}\,$

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 (talk • contribs)

feasibility of the geometric interpretation

Regarding the Geometric Interpretation: In the 2D case, 4 $a\times b$ rectangles always fit inside a $(a+b)\times (a+b)$ 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 :$2n\,$, $n^{2}\,$ or $n^{2}\,$ instead of just $n\,$. 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 + ... + x_n). 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)

I realize that replying to a ten year old comment is probably self-defeating, but I am doing this so that others don't fall into the same trap. The name "inequality of arithmetic and geometric means" is not Wikipedia-only, it is the classic name for the inequality before it became acronymized. For instance, that exact name can be found in the 2nd edition of Chrystal's Algebra: An Elementary Text-Book published in 1900. (I have not seen the 1889 edition.) Incidentally, regarding acronyms, I have seen AGM far more often than I have ever seen AM-GM. Michael Kinyon (talk) 18:51, 16 January 2024 (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 $a\geq b\geq 0,$ then setting $x=(a+b)/2$ and $y=(a-b)/2,$ so $a=x+y$ and $b=x-y,$ yields:

${\begin{aligned}(a+b)/2&={\bigl (}(x+y)+(x-y){\bigr )}/2=x\\{\sqrt {ab}}&={\sqrt {(x+y)(x-y)}}={\sqrt {x^{2}-y^{2}}}\leq x\qquad {\text{equality if and only if }}y=0,{\text{ equivalently }}a=b.\end{aligned}}$

Direct proof: This can be seen from the fact that the square of a real number is always positive, so

$0\leq (a-b)^{2}=a^{2}-2ab+b^{2}=a^{2}+2ab+b^{2}-4ab=(a+b)^{2}-4ab.$

In other words $(a+b)^{2}\geq 4ab$, with equality precisely when $a-b=0$ i.e.$a=b$. 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)

Several errors in Finance: Link to geometric asset returns

1) $g_{N}$ should have a -1 in its definition

2) The formula given relating $g_{N}$ and $a_{N}$ is, I think, an approximation not an equation.

3) I would suggest that everything in this section after "where sigma^{2} is the variance of the observed asset returns" should be deleted. It is detailed and not very important.

I'm not sure if I should make these changes myself? — Preceding unsigned comment added by Blitzer99 (talk • contribs) 13:47, 15 March 2024 (UTC)

Mistake in File:QM AM GM HM inequality visual proof.svg

There is a letter C on right side of picture. However, it must be M. Алры (talk) 21:19, 14 May 2024 (UTC)