= Solution
Use the physical root of the cubic which tends to $1/2$ as $\eta\to0$. For $0<\eta<1$, the cubic has value $1$ at $x=0$ and value $-\eta/4$ at $x=1/2$, and its derivative $6\eta x^2-(2+\eta)$ is negative throughout that interval. Thus it has exactly one root there. Implicit differentiation gives
$$
\frac{dx}{d\eta}=\frac{x-2x^3}{6\eta x^2-(2+\eta)}<0\qquad(0<x<1/2).
$$
The smallest limiting value occurs as $\eta\to1^-$, since larger $\eta$ cannot have a finite turnaround. At $\eta=1$ the equation factors:
$$
2x^3-3x+1=(x-1)(2x^2+2x-1)=0.
$$
The continuation of the collapsing root is therefore
$$
\boxed{\inf\frac{r_v}{r_m}=\frac{\sqrt3-1}{2}\simeq0.3660,\qquad
\frac{\sqrt3-1}{2}<\frac{r_v}{r_m}<\frac12\quad(0<\eta<1).}
$$
The lower endpoint is an infimum, not an attained minimum for a shell that turns around in finite time. This is the <minimum virial-radius ratio with positive cosmological constant>. The separate root $x=1$ at $\eta=1$ describes the limiting static situation and is not the collapse branch.
The smaller ratio does not mean vacuum repulsion makes turnaround easier. At fixed turnaround data, $U_m=-A-B$ is more negative than for $\Lambda=0$. If one tried $x=1/2$, the virialized energy would be $-A-B/2$, which is not negative enough to equal $U_m$. Extra contraction is necessary to increase the magnitude of the gravitational binding energy. Thus vacuum repulsion inhibits formation of a bound shell, while the shell that does form has a smaller idealized final-to-turnaround ratio under the stated energy and profile assumptions.
Back to article page
