 Solution

Minimize the Helmholtz free energy at fixed $N,T$. Its derivative isThe physical stationary point for $ϵ>0$ isTo check the minimum directly, put $x=b/\xi$. Then $F/(N{k}_{B}T)=(x-ϵ/2{)}^2-{ϵ}^2/4$, which is globally minimized at $x=ϵ/2$. The binding free energy gain per monomer is consequently $\mu =k_{B}T{ϵ}^2/4>0$.

The weak-adsorption blob picture requires $b\ll {\xi }_{\ast }\ll b\sqrt{N}$, or $2/\sqrt{N}\ll ϵ\ll 2$ in this model. If $ϵ$ is too large, the layer saturates at a monomer-scale thickness; if too small, a many-blob adsorbed layer has not formed.