Source: China Mathematical Olympiad 1992 problem1
Tags: inequalities, triangle inequality, algebra unsolved, algebra
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Let equation $x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\dots +a_1x+a_0=0$ with real coefficients satisfy $0<a_0\le a_1\le a_2\le \dots \le a_{n-1}\le 1$. Suppose that $\lambda$ ($|\lambda|>1$) is a complex root of the equation, prove that $\lambda^{n+1}=1$.
Does this problem have to do with something like?
$\lambda^{n+1}=1$
$\lambda^{n+1}-1=0$
$\frac{\lambda^{n+1}-1}{\lambda-1}=0$
$ \lambda^n=-(a_{n-1}\lambda^{n-1}+a_{n-2}\lambda^{n-2}+\cdots+a_0) $
$ \lambda^{n+1}=-(a_{n-1}\lambda^n+a_{n-2}\lambda^{n-1}+\cdots+a_0\lambda) $
$ \lambda^{n+1}-\lambda^n=-a_{n-1}\lambda^n+(a_{n-1}-a_{n-2})\lambda^{n-1}+\cdots+(a_1-a_0)\lambda+a_0 $
$ \implies |\lambda|^{n+1}=|(1-a_{n-1})\lambda^n+(a_{n-1}-a_{n-2})\lambda^{n-1}+\cdots+(a_1-a_0)\lambda+a_0| $
$ \le (1-a_{n-1})|\lambda|^n+(a_{n-1}-a_{n-2})|\lambda|^{n-1}+\cdots+(a_1-a_0)|\lambda|+a_0 $
$ \le (1-a_{n-1})|\lambda|^n+(a_{n-1}-a_{n-2})|\lambda|^n+\cdots+(a_1-a_0)|\lambda|^n+a_0|\lambda|^n = |\lambda|^n $
$ \implies |\lambda| \le 1 \implies |\lambda|=1 $. Thus, the triangle inequality holds and $ a_0=a_1=\cdots=a_{n-1}=1 \implies \lambda^{n+1}=1 $
Shouldn't it be $ |\lambda| \geq 1 $ in the question?