Talk:AM–GM inequality

message: 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)
thread_title: Untitled

message: 15:30, 24 March 2022 (UTC)
thread_title:  Proof by averaging the arithmetic mean 

message: 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)
replies: text: Looks fine to me. Thanks for the clarification. -- Jitse Niesen (talk) 04:15, 23 July 2007 (UTC)
thread_title:  Details for points 2 and 3 

message: 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)
thread_title:  Final Sentence of Proof by Induction Subsection 2 

message: 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)
thread_title:  feasibility of the geometric interpretation 

message: 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)
replies: text: 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)

text: ....now how many edges of length x1 are there?  To be continued.... Michael Hardy (talk) 17:47, 26 August 2010 (UTC)
replies: text:  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)
replies: text:  ...and so the total edge length is 2n−1(x1 + ... + xn). Michael Hardy (talk) 20:25, 26 August 2010 (UTC)
thread_title:  Problem with formula in the Geometric interpretation section 

message: 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)
replies: text: 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)
thread_title:  Article name 

message: 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 settingThis is the Discrete Fourier transform (DFT) for n=2. x = (a + b)/2 and y = (a - b)/2, so a = x + y and b = x - y, yields:
::\begin{align}
 (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{align}
: 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)
replies: text: Many thanks to User:Schmock for making this update! Quietbritishjim (talk) 16:59, 28 June 2013 (UTC)
thread_title:  Proof in lead section for n=2 

message: 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)
thread_title:  Several errors in Finance: Link to geometric asset returns 

message: There is a letter C on right side of picture. However, it must be M. Алры (talk) 21:19, 14 May 2024 (UTC)
thread_title:  Mistake in File:QM AM GM HM inequality visual proof.svg