Relation between the coefficients of $az^3+bz^2+cz+d=0$, when $\Re(z)<0$

Question

If all the three roots of $az^3+bz^2+cz+d=0$ have negative real part, and $a, b, c, d \in \mathbb{R}$, then

1. $bc>ad$
2. $bc<ad$
3. $bc=ad$
4. None of these

I'm not quite sure how to proceed with this question. Also, is it possible to solve it logically, without any lengthy calculation?

Answer 1

As the example of $$(z+1)^3=z^3+3z^2+3z+1$$ shows, the only possible options are 1 and 4. Let us show that $bc>ad$.

Since $a$, $b$, $c$ and $d$ are reals the roots of the equation have the form $$r_1=x+iy, r_2=x-iy, r_3=t$$ where $x,y,t\in\mathbb{R}$ and $x<0, t<0$. Assume for now that $a=1$ so that $$d=-r_1r_2r_3=-t(x^2+y^2)$$ while $$b=r_1+r_2+r_3=2x+t$$

and $$c=-(r_1r_2+r_1r_3+r_2r_3)=-x^2-y^2-2tx$$ which gives $$bc=(2x+t)(-x^2-y^2-2tx)=-2x(x^2+y^2)-4tx^2-t(x^2+y^2)-2t^2x$$ Since $x<0$ and $t<0$ we have $$-2x(x^2+y^2)-4tx^2-2t^2x>0$$ so that $$bc>-t(x^2+y^2)=d$$

The general case reduces to this one by looking at the equivalent equation $$z^3+\dfrac{b}{a}z^2+\dfrac{c}{a}z+\dfrac{d}{a}=0$$