$ax^3+bx^2+cx+d=0$ with coefficient $a>b>c>d>0$. Show that the magnitude of all its roots is less than $1$.

Question

Question:

$ax^3+bx^2+cx+d=0$ with coefficient $a>b>c>d>0$.
Show that the magnitude of all its roots is less than $1$.

First of all, it's easy to finish the proof if the three roots $\alpha, \beta, \gamma$ are real.
That is $\alpha^2+\beta^2+\gamma^2=(-\frac{b}{a})^2-2(\frac{c}{a})=\frac{b^2-2ac}{a^2}<\frac{b^2}{a^2}<\frac{a^2}{a^2}<1$.

But the case of $\alpha,\beta=\bar\alpha$ are conjugate complex numbers and $\gamma$ is real.
I have tried that $|\alpha|^2+|\beta|^2+|\gamma|^2=2\alpha\beta+\gamma^2$ $=2\cdot(\frac{-d}{a\gamma})+\gamma^2\geq 3\sqrt[3]{(\frac{d}{a})^2}$ by AM-GM, and it doesn't work. And also $=2\cdot(\frac{c}{a}-\gamma(\alpha+\beta))+\gamma^2$ $=2\cdot(\frac{c}{a}-\gamma\cdot(-\frac{b}{a}-\gamma))+\gamma^2$ $=3\gamma^2+\frac{2b}{a}\gamma+\frac{2c}{a}$ $\geq\cdots$ and it still cannot to be $\leq 1$. Please help, and thank you very much.

Answer 1

Another way to prove it, albeit with some more calculations, is to use The Routh–Hurwitz criterion for real cubics, which can be stated as:

The third-order polynomial $Q(x)=A x^{3} + B x^{2} + C x + D$ has all roots in the open left half-plane if and only if $A, B, C, D$ have the same sign and $B\,C \gt A\,D\,$.

Let $\,P(x) = a x^3 + b x^2 + c x + d\,$ with roots $\,x_k \,\big|_{k=1,2,3}\,$. The Möbius transformation $\,x \mapsto \frac{x+1}{x-1}\,$ maps the open left half-plane to the open unit disc, so the roots of $\,P(x)\,$ lie strictly inside the unit circle iff the roots of $\,Q(x) = (x-1)^3 P\left(\frac{x+1}{x-1}\right)\,$ lie in the left half-plane, and $Q(x)$ is:

$$ \begin{align} Q(x) &= (x-1)^3 P\left(\frac{x+1}{x-1}\right) \\ &= a(x+1)^3 + b(x+1)^2(x-1) + c(x+1)(x-1)^2 + d(x-1)^3 \\ &= \underbrace{(a + b + c + d)}_{A} x^3 + \underbrace{(3 a + b - c - 3 d)}_{B} x^2 + \underbrace{(3 a - b - c + 3 d)}_{C} x + \underbrace{a - b + c - d}_{D} \end{align} $$

$Q(x)\,$ satisfies the conditions of the Routh–Hurwitz criterion, since $\,A,B,C,D \gt 0\,$, and:

$$ \begin{align} B\,C - A\,D &= (3a + b - c - 3d)(3a - b - c + 3d) - (a+b+c+d)(a - b + c - d) \\ &= 8 \big(a(a - c) + d(b - d)\big) \\ & \gt 0 \end{align} $$

This means the roots of $\,Q(x)\,$ lie in the open left half-plane, and therefore the roots of $\,P(x)\,$ lie strictly inside the unit circle, so they all have magnitudes $\,\lt 1\,$.

Answer 2

After I read that article. I mimic the solution to do my special case problem.
Let $f(x)=ax^3+bx^2+cx+d$
Consider $g(x)=(1-x)f(x)$
$=(1-x)(ax^3+bx^2+cx+d)$
$=-ax^4+(a-b)x^3+(b-c)x^2+(c-d)x+d$
Suppose $|z|>1$ be a root of $f(x)$ and hence a root of $g(x)$, thus
$az^4=(a-b)z^3+(b-c)z^2+(c-d)z+d$
$a|z|^4=|az^4|=|(a-b)z^3+(b-c)z^2+(c-d)z+d|$
$\leq (a-b)|z|^3+(b-c)|z|^2+(c-d)|z|+d$
$<(a-b)|z|^3+(b-c)|z|^3+(c-d)|z|^3+d|z|^3$
$=a|z|^3$
contradict with $|z|>1$. So $|z|\leq 1$