Problem
We want to know for which values of $x,y,z,w$ the function $\sigma$ is positive or negative:
\begin{equation} \sigma = \frac{A}{A^2-B^2}, \end{equation}
where \begin{eqnarray} A & = & (z-w)^2(x+y)^2 + (z^2-y-2zw)^2(w^2-x)+(z^2+x-2zw)^2(w^2+y)\\ B^2 & = & 4(z^2-y-2zw)^2(z^2+x-2zw)^2(x-w^2)(w^2+y). \end{eqnarray} All of $x,y,z,w$ are positive real constants. Furthermore, we know that
\begin{eqnarray} w^2 & < & x\\ (z-w)^2 & < & w^2+y \quad \Rightarrow\\ z^2-2zw & < & y. \end{eqnarray}
My Attempt
I begin by noting that the middle term in $A$ is the only negative one: \begin{eqnarray} A & = & \underbrace{(z-w)^2(x+y)^2}_{>0} + \underbrace{(z^2-y-2zw)^2}_{>0}\underbrace{(w^2-x)}_{<0}+\underbrace{(z^2+x-2zw)^2(w^2+y)}_{>0}\\ B^2 & > & 0. \end{eqnarray} We can divide the problem into the following steps
- find for which values the middle term dominates A
- find for which values the magnitude of $|A|$ is greater/less than $|B|$.
Step 1 gives the sign of the nominator, step 2 the denominator. The combination of these steps solves the problem. I tried to solve this both by hand and with Mathematica, but I really cant seem to implement it correctly in Mathematica.