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…
Best Answer
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.