For a path of length $D$, let $r$ be its integer midpoint. Every vertex lies beyond $r$ from at least one endpoint. If both endpoints buy a link to $r$, their combined distance savings are at least $n(\lfloor D/2\rfloor-1)$. The two Nash inequalities bound this by $2\alpha$, giving $D\le 3+4\alpha/n$.

Positive edge price rules out duplicate purchases at equilibrium. A tree therefore has $n-1$ purchases, and its social cost is at most $5\alpha(n-1)+3n(n-1)$. Every connected profile has social cost at least $\alpha(n-1)+n(n-1)$; disconnected profiles have infinite cost. Since $n\ge2$, the former upper bound is strictly less than five times the latter lower bound. The optimum is positive and finite, so division proves the claim.

The conclusion is Theorem 3 of Fabrikant, Luthra, Maneva, Papadimitriou, and Shenker, On a Network Creation Game (PODC 2003, p. 350). This proof uses a midpoint argument instead of the paper's centroid argument. Tree paths, distances, and finite edge counting use mathlib.