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Source statement and correction. Mathialagan, published 2021 PDF, pp. 9--10, defines the energy using nonzero distances but states D(P,Q)≥m2n2/∣E(P,Q)∣D(P,Q)\geq m^2n^2/|E(P,Q)|. That exact inequality needs correction when PP and QQ overlap. Here ∣P∣=m|P|=m, ∣Q∣=n|Q|=n, 2≤m≤n2\leq m\leq n, and s=∣P∩Q∣s=|P\cap Q|. Let D+D_+ count positive cross-distances, while DD also counts zero if it occurs. The corrected statement is

E={(p1,q1,p2,q2)∈P×Q×P×Q:∣p1−q1∣=∣p2−q2∣>0},D≥D+≥(mn−s)2∣E∣≥m2n24∣E∣.E=\{(p_1,q_1,p_2,q_2)\in P\times Q\times P\times Q: |p_1-q_1|=|p_2-q_2|>0\}, \qquad D\geq D_+\geq\frac{(mn-s)^2}{|E|} \geq\frac{m^2n^2}{4|E|}.

Proof. For each positive distance δ\delta let eδe_\delta count its ordered pairs in P×QP\times Q. Exactly ss of the mnmn pairs have zero distance, so ∑δeδ=mn−s>0\sum_\delta e_\delta=mn-s>0. Two ordered pairs of the same positive distance specify exactly one member of EE, giving ∣E∣=∑δeδ2|E|=\sum_\delta e_\delta^2. Cauchy--Schwarz gives (mn−s)2≤D+∣E∣(mn-s)^2\leq D_+|E|. Finally s≤ms\leq m and n≥2n\geq2 imply mn−s≥mn/2mn-s\geq mn/2. This proves all the displayed inequalities.

For example, if P=QP=Q consists of two points, then D=2D=2 and ∣E∣=4|E|=4; the uncorrected right side is 44. The correction changes only an absolute constant in the later asymptotic deduction.

Dependencies and use. Only the finite Cauchy--Schwarz inequality is used. This is the last energy-to-distance step in Theorem 3.

Verification scope. Verified within the independently reviewed Theorem 3 chain, retained in the final review; this compilation-supplied overlap correction and its application belong to the living record on Theorem 3.

Bears on. Problem 661.