# Is most of the GM-AM Inequality in its codicil?

Let’s define the codicil of the Geometric Mean – Arithmetic Mean Inequality to be the statement that if the means are equal, then all the terms are equal. Then: I conjecture that most of the GM-AM Inequality is actually in its codicil, in the sense that if you can use the codicil (that is, assume the codicil as part of the hypothesis), then it is a hundred times easier to prove the AM-GM Inequality.

### Rephrased by Y.F.

Is a proof along the following lines possible? Show that as long as not all variables are equal, there is a way to make them "more equal" which decreases the difference $\mathrm{AM}-\mathrm{GM}$. Deduce somehow that "the worst case" is when all variables are equal, in which case we know that $\mathrm{AM}=\mathrm{GM}$.

### Addendum by OP Mike Jones

I thought it was obvious that although the setting of this question is real analysis (specifically, inequalities), the germane issue is proof theory. I want to explore the interrelationships between the various parts of the hypothesis and conclusion of this celebrated theorem. I would have tagged it as “proof theory” in the first place, but didn’t see “proof theory” in my cursory glance at the tags. Now, I have found it buried under the tag “logic”. I was attempting to add the “logic” tag when I got thrown into the comment box, so, could someone who knows how please add this tag to the question? Also, if the answer to the question is affirmative, then it should also be tagged with “education” and “teaching”. About 80% of learning consists of simply acquiring (deep) familiarity, and so the teacher could pose the problem to the students, or to an especially talented/eager younger student: “Prove the GM-AM Inequality, but use its codicil as part of the hypothesis this first time around.”. Indeed, that is the original motivation for my question: how to make this celebrated result much more accessible to students. (I teach mathematics in high school.) It occurred to me that perhaps a great deal, even most, of the difficulty is tied up in the codicil, precisely as user “Moron” has described: (GM = AM implies values equal) implies GM <= AM.

I found that this is too long for a comment, so I bailed out of the comment box and am putting this in as an addendum to the original question.

Oh, I see now how to do addtional tags:-)

Regards, Mike Jones American expatriate in Beijing 5.May.2011

• @Yuval: I don't understand the rephrasing either. Especially: " thus showing moreover equality holds only if all terms are equal." Can you please elaborate? – Aryabhata Apr 30 '11 at 1:45
• @Moron: Do you like my new rephrasing better? – Yuval Filmus Apr 30 '11 at 2:44
• @Yuval: If you assume the equality part, aren't you already assuming the stronger claim? – Aryabhata Apr 30 '11 at 3:20
• @Mike What is your motivation for this question? The A-G mean inequality is well known as is many of its proofs. What do you hope to gain from an answer to this question? – JavaMan Apr 30 '11 at 3:24
• @Mike: We have to use something more about the means themselves, since the codicil is symmetric in AM and GM. Do you have any thoughts at what that extra property could be? – Yuval Filmus May 4 '11 at 23:51

One way to prove the AM-GM inequality along the lines suggested by my rephrasing is as follows. Denote the variables by $x_1,\ldots,x_n$, all positive. Suppose that it is not true that all variables are equal. There are thus variables $x_i,x_j$ which are different, say $x_i < x_j$. Suppose we brought them closer together by letting $x'_i = x_i + \epsilon, x'_j = x_j - \epsilon$ for some $0 < \epsilon < x_j - x_i$. This doesn't change the arithmetic mean but increases the geometric mean: $$x'_i x'_j = (x_i + \epsilon)(x_j - \epsilon) = x_i x_j + \epsilon (x_j - x_i - \epsilon) > x_i x_j.$$ Note that any $\epsilon$ for which $x'_i \leq x'_j$ will do: such an $\epsilon$ satisfies the stronger requirement $\epsilon \leq (x_j-x_i)/2$.

Using $n-1$ such operations we can go from the original assignment to the constant assignment, in which the two means are the same. This shows that the geometric mean is always at most the arithmetic mean, with equality only if all variables have the same value.

The dual proof maintains the geometric mean while decreasing the arithmetic mean. Details left to the reader. A similar proof should also work for the harmonic mean - geometric mean inequality.

A different proof along similar lines runs as follows. Given $x_i \neq x_j$, modify the variables by averaging $x_i,x_j$. Since $$x_i x_j = \left(\frac{x_i + x_j}{2} + \frac{x_i - x_j}{2}\right) \left(\frac{x_i + x_j}{2} - \frac{x_i - x_j}{2} \right) = \left(\frac{x_i + x_j}{2}\right)^2 - \left(\frac{x_i - x_j}{2}\right)^2,$$ this operation increases the geometric mean while not changing the arithmetic mean. Intuitively, it's obvious that if we apply these operations successively then each variable tends to the arithmetic mean (this can be proven formally using a potential function argument). Since the geometric mean is continuous, the AM-GM inequality follows.

Affirmative.

It is indeed very easy to prove the “main part” of the celebrated GM-AM Inequality from its codicil (and by appealing to the continuity of the functions involved).

Note that if the GM-AM Inequality holds for positive integers, then it is trivially easy to show that it also holds for positive rational numbers (-for if AM < GM, then w*AM < w*GM, where w is the product of the denominators. w can then be pulled into each expression, yielding positive integers for which the GM-AM Inequality fails, a contradiction). Furthermore, if the GM-AM Inequality holds for all positive rational numbers, the by continuity (of the functions involved), it holds for all positive real numbers. Hence, the whole edifice can be made to rest on whether the GM-AM Inequality holds for positive integers. But if we can assume its codicil, proving that it holds for positive integers is very easy to do, as shown below.

Codicil Axiom. If n is a positive integer and x(1), …, x(n) are positive real numbers such that GM(x(1),…,x(n)) = AM(x(1), …,x(n)), then x(1) = x(2) = … = x(n).

Theorem. If n is a positive integer and x(1), …, x(n) are positive integers, then GM(x(1),…,x(n)) <= AM(x(1),…,x(n)).

Proof: By contradiction. Suppose that there exists a positive integer n and positive integers x(1),…, x(n) such that AM(x(1),…,x(n)) < GM(x(1),…,x(n)). Since it is trivially easy to establish the GM-AM Inequality for n = 1 and n = 2 we can assume, wlog, that n > = 3. Furthermore, in any such case, we can assume, wlog, that x(1) <= … <= x(n). Let M(0) be the set {(x(1),…,x(n)) | x(1), …, x(n) are positive integers such that x(1) <= … <= x(n) and AM(x(1),…,x(n)) < GM(x(1),…,x(n)). By hypothesis, M(0) is non-empty. Since its members are positive-integer-based, M(0) has a (lexicographically) “smallest” element. We want to find it. To that end, we squeeze down on it by means of the following definition. For each positive integer k <= n, let M(k) be the set {(x(1),…,x(n) | (x(1),…,x(n))  M(k – 1) and for each (y(1),…y(n))  M(k – 1), x(k) <= y(k)). Then M(n) contains exactly one element, which we will denote by (x(1),…,x(n)). (It is the lexicographically smallest element of M(0) that we were looking for.) And, of course, it is the case that x(1) < x(n). (It is important to bear in mind throughout that x(1) <= … <= x(n).)

Case 1. x(1) > 1. Change the value of x(1) to x(1) – 1 in a manner that is monotone and continuous. Then the relation AM < GM will have changed to GM <= AM by, or at, the time x(1) – 1 is reached (by the minimality of x(1), and by the fact that x(1) – 1 is a positive integer, since x(1) is a positive integer > 1). But for this to happen, the relation GM = AM must (by continuity of the functions involved) be the case somewhere along the way. However, it cannot happen in the open interval (x(1) – 1,x(1)), because that is a non-integer value, and, by the Codicil Axiom, x(2), …, x(n), which are all integers (with n > 2) would have to be equal to it, which would be a contradiction. Thus equality of means must occur for x(1) – 1. That is, GM(x(1) – 1,…,x(n)) = AM(x(1) – 1,…,x(n)). Then by the Codicil Axiom, x(n) = x(1) – 1. Then x(n) < x(1), a contradiction.

Case 2. x(1) = 1.

Case 2.1. x(n) > 2. Change the value of x(n) to x(n) – 1 in a manner that is monotone and continuous. Then the relation AM < GM will have changed to GM < AM by, or at, the time x(n) – 1 is reached. But for this to happen, the relation GM = AM must (by the continuity of the functions involved) be the case somewhere along the way. However, it cannot happen in the open interval (x(n) – 1,x(n)), because that is a non-integer value, and, by the Codicil Axiom, x(1), …, x(n – 1) (with n > 2), which are all integers, would have to be equal to it, which would be a contradiction. Thus equality of means must occur for x(n) – 1. That is, GM(x(1),…,x(n – 1),x(n) – 1) = AM(x(1),…x(n – 1),x(n) – 1). Then by the Codicil Axiom, x(1) = x(n) – 1. But x(n) – 1 > 1, since x(n) > 2. Thus x(1) > 1, a contradiction.

Case 2.2. x(n) = 2 and x(n – 1) = 2. Moving x(n) from 2 to 1 forces x(n – 1) to be 1, a contradiction.

Case 2.3. x(n) = 2 and x(n – 1) = 1. In this case, remembering that x(1) <= … <= x(n – 1) <= x(n), the relation GM < AM can easily be established by direct calculation.

This covers all the cases, thereby establishing the result.

(end of proof)

Now, assuming I haven’t blundered somewhere in this attempt. I ask you: Isn’t this a hundred times easier than any of the proofs of the full result?

Regards, Mike Jones