 Solution

The Kullback-Leibler divergence of a product measure is the sum of the individual divergences, by expanding the logarithm of the likelihood ratio and taking its expected value. Put $q=1/2-t/\sqrt{n}$ and define $W(S)$ to be the unordered pairs lying within one group, while $C(S)$ is the set of unordered pairs crossing between groups. ThenThus choosing the printed $\partial S$ to mean $W(S)$ gives its order of the summands. If $\partial S$ denotes the usual graph cut $C(S)$, their two orders must be interchanged. Each unordered edge is counted once.

The inequality $\log x\le x-1$ givesSince $1/3<q<1/2$, each changed-edge divergence is at most $9t^2/(2n)$. Summing over at most $n(n-1)/2$ pairs gives $D_{\mathrm{KL}}(P_S\Vert P_{S^{\prime }})\le 9n{t}^2/4$. This is the information scale needed for Fano's inequality.