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I'm currently reading the following paper: http://arxiv.org/abs/1209.0612 and got stuck on Proposition 3.1 (2).

The claim translated to polynomials is the following:

Assume $n\geq 3, c\geq 1, d\geq 1$ are natural numbers such that $c²+d²-(n-1)cd<0$. Show that $(n³-n+1)c²+(n+1)d²-(n²+n-1)cd>1$.

Anyone an idea to solve this?

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For $n=3$, the equation $c^2+d^2 - 2 cd < 0$ has no solutions. –  copper.hat Sep 13 '12 at 16:36
@copper.hat - the question asks to assume there is a solution, it does not claim that all $n\geq 3$ have solution –  Belgi Sep 13 '12 at 16:42
@copper.hat - that's right and this has a representation-theoretic interpretation. –  Julian Kuelshammer Sep 13 '12 at 16:46
@Belgi: This is true, of course, but false presumptions can often lead to strange results. I just checked the simplest boundary case and the presumption was false, so, being lazy, I stopped there... –  copper.hat Sep 13 '12 at 16:50
As it turns out, the condition $c^2+d^2-(n-1)cd<0$ isn't even needed. (see Answer below). –  Hagen von Eitzen Sep 13 '12 at 17:00

1 Answer 1

up vote 4 down vote accepted

Let $c,d$ be positive real numbers and $n>1$ (esp., $n^3-n+1>0$). By the arithmetic-geometric inequality $$(n^3-n+1)c^2+(n+1)d^2\ge 2\cdot\sqrt{(n^3-n+1)(n+1)}\cdot c d.$$ One checks by multiplying out that $$4(n^3-n+1)(n+1)=(n^2+n-1)^2+3+3n^2(n^2-1)+2n(n^2+1),$$ hence $2\cdot\sqrt{(n^3-n+1)(n+1)}>n^2+n-1$ and finally $$(n^3-n+1)c^2+(n+1)d^2>(n^2+n-1) cd. $$

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Thank you very much. –  Julian Kuelshammer Sep 13 '12 at 16:45

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