$\omega$ satisfies $a \omega^3 + b \omega^2 + c \omega + d = 0$, Prove that $ |\omega| \leq \max( \frac{b}{a}, \frac{c}{b}, \frac{d}{c})$

Question

If $\omega$ is a complex number that satisfies $a \omega^3 + b \omega^2 + c \omega + d = 0$, where $a,b,c,d$ are positive real numbers, prove that $|\omega| \leq \max( \frac{b}{a}, \frac{c}{b}, \frac{d}{c})$

This just seems so simple and elegant yet I can't wrap my head around. Was trying to tackle from vieta's formula but nothing fruitful yet.

Another idea was to make the polynomial $\omega^3 + \frac{b}{a} \omega^2 + \frac{c}{b} \omega + \frac{d}{c} = 0$, and prove some contradiction if $\omega > \max( \frac{b}{a}, \frac{c}{b}, \frac{d}{c})$. It might be closer to what we want.

Answer 1

Let $r=\max( \frac{b}{a}, \frac{c}{b}, \frac{d}{c})$. To simplify the problem, use the substitution $\omega =rz$, $$ ar^3z^3+br^2z^2+crz+d=0,\\ az^3+\frac{b}{r} z^2+ \frac{c}{r^2} z+\frac{d}{r^3}=0,\\ z^3+B z^2+ C z+D=0, D\leq C\le B\le 1. $$ Now, consider $$ f(z)=(1-z)(z^3+B z^2+ C z+D)+z^4=(1-B)z^3+(B-C)z^2+(C-D)z+D. $$ For $|z|=1$, $f(z)\leq |1-B|+|B-C|+|C-D|+D=1$, so $|z^3f(1/z)|\leq 1\times 1=1$.

By maximum modulus principle, $|z^3f(1/z)|\le 1$ for all $|z|\le 1$. That means for $|z|>1$, $|f(z)/z^3|\le 1$, $|f(z)|\le |z|^3$ Therefore, $$ |(1-z)(z^3+B z^2+ C z+D)|=|f(z)-z^4|\ge -|f(z)|+|z|^4\\ \ge |z|^4-|z^3|=|z|^3(1-|z|)>0 $$ for $|z|>1$, so for $(1-z)(z^3+B z^2+ C z+D)=0$, $|z|\leq 1$. As a result, $$ |\omega|\le |r||z|\le r. $$ For more information, see this document.

Answer 2

Here is an answer that takes advantage of the simplified context given by the four first lines of the answer by @Jethro, switching from there to a simple geometrical proof.

Let us assume indeed that

$$z^3+B z^2+ C z+D=0, \ \ \ D\leq C\le B\le 1 \ \tag{1}$$

where we have to show that $|z| \leq 1$.

Let us consider the case of a complex root

$$z:=\rho e^{i \theta}\tag{2}$$

(we can assume that $0<\theta<\pi$ possibly up to the consideration of the complex conjugate root).

Let us show that necessarily $\rho\leq 1$.

Assume on the contrary that $\rho>1$, in order to obtain a contradiction.

Consider for that the quadrilateral with vertices

$$P:=0, \ \ Q:=D, \ \ R:=D+Cz, \ \ S:=D+Cz+Bz^2$$

whose sidelengths $D,Cr,B\rho^2,\rho^3$ are ranked in the following order (as a consequence of (1) and the fact that $r>1$):

$$D < Cr < B\rho^2 < \rho^3$$

i.e., $PQRS$ is a quadrilateral with equal opposite angles $a=\pi-\alpha$ (in vertices $Q$ and $S$) and increasing sidelengths. But this is impossible (See the answer to my recent question How to prove that no quadrilateral exist with three equal angles and increasing sidelengths).

Remark : a different computation shows that one gets as well a contradiction for a real root $z=-\rho$ with $\rho>1$.