I have to prove, that polynomial:
$\ x^n + a_{n-1}x^{n-1}+...+a_2x^2+a_1x+a_0=0$
, where $ a_i= [-1,0,1] $ and $ i=0,1,.....,n $
doesn't have any solutions in the set
$$
\begin{aligned}
( -\infty,-2) \cup (2, \infty)
\end{aligned}
$$
We know, that $ a_n=1$, so we know, that $ x_1+x_2+....+x_n=-2$.
It's all I have though, any
suggestions, what can I do in next step?
Let $f(x)=0$, then \begin{align} 0=&|x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0|\\ \geq&|x|^n-|a_{n-1}||x|^{n-1}-\cdots-|a_1||x^1|-|a_0| \end{align} Write it as \begin{align} |x|^n\leq & |a_{n-1}||x|^{n-1}+\cdots+|a_1||x^1|+|a_0|\\ \leq & |x|^{n-1}+|x|^{n-2}+\cdots+|x|+1\\ = & \frac{|x|^{n}-1}{|x|-1} \end{align} If $|x|>2$, we get $|x|-1>1$ and then $$ |x|^n\leq |x|^n-1. $$ It is impossible.