= Uniqueness of positive hexagon equilibria with straddling cross-couplings
{title2=$(b-a)(c-a)<0$}
At a positive equilibrium set $x_i=r_i^2$ and $p=r_1r_2r_3$. The equations imply $a x_i+b x_{i+1}+c x_{i+2}-\alpha p/x_i=\mu$. Suppose $b>a>c$ and relabel cyclically so $x_1$ is maximal. If $x_2\geq x_3$, subtracting equations one and two makes the cubic side positive unless all are equal, whereas $\alpha p(1/x_1-1/x_2)$ is nonpositive. If $x_2\leq x_3$, subtracting equations two and three gives the opposite sign contradiction. Interchanging the orientation handles $c>a>b$. Thus all positive equilibria have equal amplitudes. With $\alpha\ne0$, a state with exactly two nonzero amplitudes cannot be an equilibrium because it immediately forces the third.
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