Elementary bound theorem of a monic real polynomial

Question

An elementary bound theorem on the roots of a real monic polynomial states that $$M := \operatorname{max} (1, |a_0| + \cdots + |a_{n-1}|) := \operatorname{max} (1, B)$$ is an upper and lower ($-M$) bound for the polynomial's roots. The proofs I know are either ugly (using synthetic substitution), or overkill ("follows from (a corollary of) Rouche's theorem").

To show the case when $M=1$ is easy. I struggle to find a nice, elementary proof when $M=B$. Searching MSE's database (as well as googling), I couldn't find one.

Answer 1

For $|x|\ge1$, we have $$ |a_0+a_1x+\cdots+a_{n-1}x^{n-1}| < |a_0| + |a_1| |x| + \cdots + |a_{n-1}| |x|^{n-1} \le B|x|^{n-1}. $$ Therefore when $|x|>B$ as well, $$ |a_0+a_1x+\cdots+a_{n-1}x^{n-1} + x^n| > |x|^n - B|x|^{n-1} > 0, $$ and so the polynomial can't vanish.

Answer 2

Let $\alpha$ be a root of the polynomial $p(x)=x^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$, by definition, we have: $$\alpha^n=-(a_{n-1}\alpha^{n-1}+\dots+a_1\alpha+a_0)$$ so we have: $$\lvert \alpha\rvert^n =\lvert a_{n-1}\alpha^{n-1} +\dots+ a_1\alpha+ a_0\rvert $$ Now if $\lvert\alpha\rvert>B$, $\lvert \alpha\rvert^n >B^n$. However, if $B\ge 1$, we have: \begin{align*} \lvert a_{n-1}\alpha^{n-1} +\dots+ a_1\alpha+ a_0\rvert &\le\lvert a_{n-1}\rvert\lvert\alpha^{n-1}\rvert+\dots+\lvert a_1\rvert\lvert\alpha\rvert + \lvert a_0\rvert \\ &\le (\lvert a_{n-1}\rvert\lvert B^{n-1}\rvert+\dots+\lvert a_1\rvert\lvert B \rvert + \lvert a_0\rvert \\ &\le (\lvert a_{n-1}\rvert+\dots+\lvert a_1\rvert+\lvert a_0\rvert)B^{n-1}=B^n \end{align*} Thus, $\lvert \alpha\rvert^n$ should be both $>B^n$ and $\le B^n$. Contradiction.