Prove that $a_3 \lambda^{3} + a_2 \lambda^{2} + a_1 \lambda + a_0 = 0$ has three real roots

Question

I'm trying to prove that the cubic equation

$a_3 \lambda^{3} + a_2 \lambda^{2} + a_1 \lambda + a_0 = 0$

has three real roots. The coefficients are

$a_3 = - 1 - \sigma - \tau - \chi$

$a_2 = -2 (\sigma + \tau + \sigma \tau)$

$a_1 = \chi(\tau + \sigma) + (\chi - 3) \sigma \tau$

$a_0 = 2 \sigma \tau \chi$

where each of $\sigma$, $\tau$ and $\chi$ are greater than zero. Applying the Rule of Descartes indicates that the polynomial has one real positive root and zero or two real negative roots. I've played around with forming the discriminant for the cubic, and a 3-D contour plot in Mathematica suggests that the discriminant is positive for $\sigma, \tau, \chi > 0$ (indicating three real roots). However, I've hit a wall trying to come up with a formal proof.

One thing of note is that the polynomial coefficients are unchanged on interchange of $\sigma$ and $\tau$, and so the discriminant is a symmetric function of $\sigma$ and $\tau$. That may be helpful...

Answer 1

We have \begin{align*} f(-\sigma) &= \sigma (\sigma + \chi)(\sigma - 1)(\sigma - \tau), \\ f(-\tau) &= -\tau (\tau + \chi)(\tau - 1)(\sigma - \tau), \\ f(-1) &= (\chi + 1)(\tau - 1)(\sigma - 1). \end{align*}

Thus, we have $$f(-\sigma)f(-\tau)f(-1) = - \sigma (\sigma + \chi) \tau (\tau + \chi) (\chi + 1)(\sigma - 1)^2(\sigma - \tau)^2(\tau -1)^2 \le 0.$$

Thus, at least one of $f(-\sigma), f(-\tau), f(-1)$ is less than or equal to zero. Note that $f(0) > 0$. Thus, there is a negative root.

Note that $f(0) > 0$ and $f(\infty) = -\infty$. Thus, there is a positive root.

Thus, the cubic has three real roots.

Answer 2

For your polynomial $f$, having three real roots is equivalent to the following conditions: a) the derivative $f'$ has two real roots, say $x_1<x_2$, and b) $f(x_1)f(x_2)<0$. Since $f'$ is of degree 2, these conditions are easy to verify.

Answer 3

Based on the prompting by Alexandre (thanks!), I think I've come up with a proof. The cubic discriminant is

$\Delta = 18 a_3 a_2 a_1 a_0 – 4 a_2^3 a_0 + a_2^2 a_1^2 – 4 a_3 a_1^3 – 27a_3^2 a_0^2$

...and the proof involves demonstrating that $\Delta$ is non-negative for $\sigma, \tau, \chi > 0$.

Substituting in the definitions of $a_3,\ldots,a_0$, $\Delta$ takes the form

$\Delta = 4(\tau -1)^4 (1+\chi) (\sigma+\chi) D$

where

$D = \tau^{3} \chi + 3 \sigma^2 \tau \left[ -3\tau(\tau+1) + \chi + \tau \chi (\tau - 7)\right] + \sigma^3 \left\{ 9 \tau \left[ 1 + \tau (\tau-1) \right] + (\tau+1)^3 \chi \right\} + 3 \sigma \tau^2 \left[ \chi + \tau(\chi+3)\right].$

Since the leading factors in the above expression for $\Delta$ are non-negative for $\tau, \sigma, \chi > 0$, we can now focus just on the non-negativity of $D$. First, we note that $D$ is linear in $\chi$:

$D = d_{0} + d_{1} \chi$

where

$d_{0} = 9 \sigma \tau \left\{ \tau^2 - \sigma \tau \left[ \tau + 1 \right] + \sigma^2 \left[ 1 + \tau(\tau-1) \right] \right\}$,

$d_{1} = \tau^{3} + 3 \sigma \tau^2 (\tau+1) + \sigma^2 (\tau + 1)^3 + 3 \sigma^2 \tau \left[ 1 + \tau(\tau-7) \right]. $

Suppose both $d_{0}$ and $d_{1}$ are non-negative for all $\sigma, \tau > 0$; then $D$ will be non-negative for all $\chi > 0$. The proof that $d_{0}$ is non-negative is trivial. For $d_{1}$, consider the two contours

$\frac{\partial d_{1}}{\partial \sigma} = 0, \qquad \frac{\partial d_{1}}{\partial \tau} = 0$

in the $\sigma, \tau >0$ half-plane. These contours are closed curves that are mirror images through the line $\sigma = \tau$ (a consequence of the symmetry of $d_{1}$ w.r.t. $\sigma$ and $\tau$). The contours intersect twice along this line:

• The intersection at $\sigma = \tau = \sqrt{13}-3$ is a saddle point of $d_{1}$, and at this point $d_{1} = 8 (403 \sqrt{13} - 1453) \approx 0.297$.
• The intersection at $\sigma = \tau = 1$ is a local minimum of $d_{1}$, and at this point $d_{1} = 0$.

These are the only extrema of $d_{1}$ in the half-plane. Therefore, over all $\sigma, \tau > 0$ the minimum value of $d_{1}$ is zero, and $d_{1}$ is confirmed to be non-negative.

This completes my proof. It's not particularly elegant, and I'd appreciate hearing any shortcuts that others can spot.