Existence of positive real $R$ such that $|f(z)| > M$ for all $|z| > R$ Exercise: If $f(z)$ is a polynomial of degree $n \geq 1$ and $M$ is an arbitrary positive real number, show that there exists a positive real number $R$ such that $|f(z)| > M$ for all $|z| > R$.
Claim: For $R > \frac{M}{\min_{0 \leq j \leq n} \{ |a_{j}| \}}$ with $R, M \in \mathbb{R}_{\geq 0}$ and $n \geq 1$ we have $|f(z)| > M$ for all $|z| > R$.
Proof of claim: $$R > \frac{M}{\min_{0 \leq j \leq n} \{ |a_{j}| \}} \implies \min_{0 \leq j \leq n} \{ |a_{j}| \} \cdot R > M \implies \min_{0 \leq j \leq n} \{ |a_{j}| \} \sum_{j=0}^{n} R^{n} > M \\ \implies \sum_{j=0}^{n} |a_{j}| R^{n} > M \implies \sum_{j=0}^{n} |a_{j}| |z|^{n} > M.$$ But now I can't use the triangle inequality to conclude that for a polynomial $f(z) = \sum_{j=0}^{n} a_{j}z^{j}$ of degree $n \geq 1$ to conclude that $|f(z)|  > M$ for all such $z$.
So is this claim true at all or how to find a $R$ such that we know that $|f(z)| > M$ for all $|z| > R$?
 A: Here's a direct proof. Let
$$f(x)=\sum_{m=0}^n a_m x^m$$
and let $A=\max\{|a_0|,|a_2|,...,|a_{n}|\}$. Then for
$$|z|>R=\max\left\{2,\frac{5(A+2^{-n}M)}{|a_n|}\right\}$$
we have
$$|f(z)|=|\sum_{m=0}^n a_m z^m|\geq |a_nz^n|-\sum_{m=0}^{n-1} |a_m z^m|\geq |a_nz^n|-A\sum_{m=0}^{n-1} | z^m|$$
$$= |a_n||z|^n-A\frac{|z|^n-|z|}{|z|-1}=|z|^n\left(|a_n|+A\frac{\frac{1}{|z|^{n-1}}-1}{|z|-1}\right)$$
$$>|z|^n\left(|a_n|-A\frac{1}{|z|-\frac{|z|}{2}}\right)=|z|^n\left(|a_n|-A\frac{1}{|z|-\frac{|z|}{2}}\right)$$
$$=|z|^n\left(|a_n|-2A\frac{1}{|z|}\right)\geq 2^{n-1}R\left(|a_n|-\frac{2A}{R}\right) $$
Case 1: If $R=2$ then
$$2\geq \frac{5(A+2^{-n}M)}{|a_n|}$$
$$|a_n|=\frac{5}{2}(A+2^{-n}M)>\frac{5}{2}A$$
which gives us
$$|a_n|-\frac{2A}{R}=|a_n|-A>\frac{5}{2}A-A=\frac{3}{2}A$$
Alternatively, we also have
$$2\geq \frac{5(A+2^{-n}M)}{|a_n|}>\frac{5M}{2^n|a_n|}\geq \frac{5M}{2^nA}$$
$$A>\frac{5}{2}\cdot \frac{M}{2^n}$$
The total value is then
$$2^{n-1}R\left(|a_n|-\frac{2A}{R}\right)>2^{n}\cdot \frac{3}{2}A>\frac{3}{2}\cdot \frac{5}{2}M>M$$
Case 2: If $R=\frac{5(A+2^{-n}M)}{|a_n|}$ then
$$2^{n-1}R\left(|a_n|-\frac{2A}{R}\right)>2^{n}\frac{A+2^{-n}M}{|a_n|}\left(|a_n|-\frac{A}{\frac{A+2^{-n}M}{|a_n|}}\right)$$
$$=2^n(A+2^{-n}M)\left(\frac{A+2^{-n}M}{A+2^{-n}M}-\frac{A}{A+2^{-n}M}\right)$$
$$=2^n (A+2^{-n}M)\left(\frac{2^{-n}M}{A+2^{-n}M}\right)=2^n \cdot 2^{-n}M=M$$
Either way, we conclude $|f(z)|>M$.
A: If $f(z)=\sum\limits_{k=0}^{n} a_kz^{k}$ with $a_n \neq 0$  then $f(z)/z^{n}=(\sum\limits_{k=0}^{n} a_kz^{k-n}) \to a_n$ as $|z| \to \infty$. Hence, there exists $R$ such that  $|f(z)/z^{n}| >\frac {|a_n|} 2$ and $|f(z)| >R^{n} \frac {|a_n|} 2>M$ if  $|z|>R$.
