# Generalizing an approach to proving AMGM

This problem is Exercise 5.5.30 of "The Art and Craft of Problem Solving" by Paul Zeitz.

The problem asks to use the identity $$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$ to prove the AMGM inequality. I have done this by noting that the right side is equal to $$(a+b+c)\frac{1}{2}((a-b)^2 + (b-c)^2 + (a-c)^2)$$ and then substituting $a=x^{1/3}$, $b=y^{1/3}$, $c=z^{1/3}$.

The problem then asks "Does this method generalize?", and I'm having trouble finding one. Is there a factorization of $x_1^n + x_2^n + \dots x_n^n - nx_1x_2\cdots x_n$ that can prove AMGM for $n$ variables? Or perhaps using some other generalization of the given identity?

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– copper.hat May 15 '12 at 17:53
@shirker Why do you assume that the answer to "Does this method generalize?" is in affirmative? – Sasha May 15 '12 at 18:29
The analogous polynomials for $n=4$ and $n=5$ are irreducible, so factoring doesn't do it. – David Speyer May 15 '12 at 18:37
@Sasha That's a good point, I did assume that. I think David's observation suggests the answer may be no. – shirker May 15 '12 at 19:53
Instead of going so far (generalizing), why don't you try for $n=4$ and share your answer – Kirthi Raman May 18 '12 at 20:10