TITLE: Inequality for polynomial value from Rudin
QUESTION [5 upvotes]: I can't follow a proof in Rudin's book. If 
$$
P(z)=z^n+a_{n-1}z^{n-1}+ ... +a_0
$$ 
is a polynomial with complex coefficients, then
$$
r>1+|a_{n-1}|+...+2|a_0|
$$
implies $|P(r)|>|P(0)|$.

REPLY [6 votes]: Notice that the assumption $|z|>1+2|a_0|+\ldots+|a_{n-1}|$ implies that $|z|\geq1$ and so
\begin{aligned}
|P(z)| & \geq |z|^n-\big(|a_{n-1}||z|^{n-1}+\ldots+|a_1||z|+|a_0|\big)  \\
&\geq |z|^n-|z|^{n-1}\big(|a_{n-1}|+\ldots+|a_1|+|a_0|\big)\\
&\geq |z|^{n-1}\Big(|z|-\big(|a_{n-1}|+\ldots+|a_1|+|a_0|\big)\Big)\geq |a_0|
\end{aligned}