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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

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    @Yuval: I don't understand the rephrasing either. Especially: _" thus showing moreover equality holds only if all terms are equal."_ Can you please elaborate?2011-04-30
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    @Moron: Do you like my new rephrasing better?2011-04-30
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    @Yuval: If you assume the equality part, aren't you already assuming the stronger claim?2011-04-30
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    @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?2011-04-30
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    @Moron: See my answer for one interpretation which is relatively close to the original text of the question, and even closer to my rephrasing.2011-04-30
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    @Yuval: Why don't you edit the question with more details? The only interpretation I can think of currently is $(AM = GM \implies \text{numbers are equal}) \implies AM \ge GM$. (Yes, I read your answer, but I still don't get your interpretation).2011-04-30
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    @Moron: How about now? I also added a new proof sketch which shows an (arguably) more naive approach which also works.2011-04-30
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    @Yuval: I am not really sure if that is what Mike intended. The "codicil" is not part of the hypothesis...2011-04-30
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    @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?2011-05-04

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