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This question grew out of my recent job interview. Since the interviewers were math professors, I had a hard time searching for interesting elementary theorems in case I got asked for one. I thought topics such as the Banach-Tarski paradox, Godel's theorems, the Mandelbrot set, the Brouwer Fixed Point Theorem, etc were well-known and wouldn't do the job. However, after a cursory search, I found some to my taste:

1.Marden's theorem (or here)(It is not Marsden.) Gauss–Lucas theorem

2.The identity $\int_{0}^1 \frac{x^4(1-x)^4}{1+ x^4} dx = \frac{22}{7}- \pi$

So, my question here is an invitation to expand the list (of theorems that would get an interviewee accepted).

To recap, my criteria for selection are

  1. Not widely known,
  2. Elementary- understandable to a first year grad student, and
  3. Interesting-i.e. MOtizens, assuming they are the audience, will be delighted to have come across it.

Thank you.

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Why did you expect to be asked to state interesting elementary theorems at your job interview? It's not like a thesis defense. You tell them what you know, not what they would like you to know. –  Spiro Karigiannis Jan 3 '11 at 13:44
I am a student there and the faculty wanted to recruit tutors for freshmen/sophomores. But their selection criteria was not only ability to do the tutoring but they also wanted to know whether the applicant wants to continue a career in Mathematics, how he can entice/encourage his juniors to go on studying Math, what Math field he delights in, etc. –  Unknown Jan 3 '11 at 13:57
Even granting the somewhat strange motivation of this question: isn't it self-defeating? Whatever gets posted here will become much more widely known. Moreover, I find it somehow dishonest to get examples of one's mathematical knowledge and enthusiasm from others on the internet. I am voting to close. –  Pete L. Clark Jan 3 '11 at 14:08
I agree with Pete but could not find a good way to say it. It seems to me that in light of your interview you should be trying to figure out how to become a better candidate rather than finding ways to trick the hiring committee, or whatever, into thinking you are a better candidate than you are? –  Qiaochu Yuan Jan 3 '11 at 14:25
+1 simply for pointing out Marden's theorem. –  Mitch Harris Jan 3 '11 at 16:58
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closed as off topic by Pete L. Clark, Qiaochu Yuan, Mariano Suárez-Alvarez, Andrew Stacey, Andy Putman Jan 3 '11 at 15:21

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1 Answer

The AM-GM inequality is implied by an identity.

For a function $f=f(x_1,x_2,\dots,x_n):\mathbb R^n\to\mathbb R$ let $Pf(x_1,x_2,\dots,x_n)$ denote the sum of $f$ over the $n!$ quantities that result from all possible $n!$ permutations of the $x_i$. Then $$\frac{x_1^n+x_2^n+\dots+x_n^n}{n}-x_1x_2\dots x_n=\frac{1}{2\ n!}(\phi_1+\phi_2+ \dots \phi_n),$$ where $$\phi_k=P[(x_1^{n-k}-x_2^{n-k})(x_1-x_2)x_3x_4\dots x_{k+1}]$$ $$=P[(x_1-x_2)^2(x_1^{n-k-1}+\dots x_2^{n-k-1})x_3x_4\dots x_{k+1}]\geq0.$$ The proof dates back to Hurwitz.

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Amazing. Thanks. I am checking for $n= 2, 3,\ldots$. –  Unknown Jan 3 '11 at 14:33
Am I hallucinating here or is anyone else reminded of the usual constructions of contracting homotopies? –  darij grinberg Jan 3 '11 at 15:12
What do you mean Darij? –  J.C. Ottem Jan 3 '11 at 16:28
OK, I was hallucinating. In homological algebra there is no need to symmetrize for obtaining telescoping sums. –  darij grinberg Jan 3 '11 at 17:35
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