$\omega$ 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})$ 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. 

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,
$$
|\omega|\le |r||z|\le r.
$$
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