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Differentiating the shell energy, or using its force, gives $\ddot{r}=-GM/r^2+\Lambda r/3$. At a genuine finite turnaround, $\dot{r}=0$ and the acceleration must be inward. ThusEquivalently the turnaround radius is smaller than $r_s=(3GM/\Lambda {)}^{1/3}$. The effective potential $V(r)=-GM/r-\Lambda r^2/6$ has its maximum at $r_s$ and second derivative $V^{\prime \prime }(r_s)=-\Lambda <0$. An initially expanding shell on the inner branch can turn around when its energy lies below this maximum; energy above it allows expansion over the barrier. The local inequality on $\eta$ is the turnaround bound with positive cosmological constant.

At $\eta =1$ the force vanishes at an unstable static radius. A shell approaching the maximum-energy separatrix reaches this radius only asymptotically, not at a finite expansion-to-collapse maximum. If it arrived there with zero velocity at finite time, uniqueness of the smooth radial equation would force it to have been the static solution already. Hence the inequality is strict for the turnaround used in the collapse calculation.