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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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    Try http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#Generalizations.2012-05-15
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    @shirker Why do you assume that the answer to "_Does this method generalize?_" is in affirmative?2012-05-15
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    The analogous polynomials for $n=4$ and $n=5$ are irreducible, so factoring doesn't do it.2012-05-15
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    @Sasha That's a good point, I did assume that. I think David's observation suggests the answer may be no.2012-05-15
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    Instead of going so far (generalizing), why don't you try for $n=4$ and share your answer2012-05-18

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