The question (a complex variables qual question from a few years back) asks:
Suppose that $$p(z) = a_0 + a_1z + \cdots + a_{n-1}z^{n-1} + z^n$$ is a monic polynomial with complex coefficients. If $r$ is a root of $p$, prove that
$$|r| < \max\{1, |a_0| + \cdots + |a_{n-1}|\}$$
I proved this fact in two cases, one of which used Rouché's Theorem, but since I had to resort to more trivial algebra for the second case want to make sure that my proof isn't more complicated than it needs to be:
Call $A:=\sum_{i=0,\ldots,n}|a_i|$.
(1) (Case A < 1): then $$|p(z) - z^n|\leq A$$ on the circle $|z| = 1$ by the Triangle Inequality and $|z^n| = 1$ trivially on the same circle. Since both functions are entire and $z^n$ only vanishes at the origin Rouché applies, whence $z^n$ and $(p(z) - z^n) + z^n = p(z)$ both have $n$ zeros within the unit disc. If $|r| < 1$ then certainly $|r| < \max\{1, A\}$.
(2) (Case A>= 1): For sake of argument, suppose that $r$ is a root of $p$ such that $$r > \max\{1, |a_0| + \cdots + |a_{n-1}|\} = A$$ Then $$|p(r)| = 0 = |a_0 + a_1r + \cdots +r^n|$$ One can apply the Triangle Inequality to obtain $$0 = |p(r)| \geq |r||a_1 + a_2r + \cdots + a_{n-1}r^{n-2} + r^{n-1}| - |a_0|$$ whence $$|a_0|>|a_0|/|r| \geq |a_1 + a_2r + \cdots + a_{n-1}r^{n-2} + r^{n-1}|$$ Repeated application of the same step results in $$|a_0| + \cdots + |a_{n-2}| > |a_{n-1} + r|$$ whence, by one more application of the Triangle Inequality, $$A > |r|$$ which is a contradiction of our selection criterion for $r$.
I'd really appreciate your insight. Thanks!
If $|r|<1$, then there is nothing to prove. Thus, let us assume $|r|\ge 1$. Since $$ a_0 + a_1r + \cdots + a_{n-1}r^{n-1} + r^n =0 $$ then $$ r = -\frac{a_0}{r^{n-1}} - \frac{a_1}{r^{n-2}} - \cdots - a_{n-1}. $$ By taking absolute values and using the triangular inequality, we obtain $$ |r| \le \frac{a_0}{|r|^{n-1}} + \frac{a_1}{|r|^{n-2}} + \cdots + |a_{n-1}|. $$ Finally, using that $|r|\ge 1$ (so that $\frac{1}{|r|}\le 1$), we obtain the required upper bound $$ |r| \le |a_0| + \cdots + |a_{n-1}|. $$