Let count the triangles of the binomial random graph . Then
Two distinct triangle indicators are dependent only when the triangles share an edge. Hence their ordered Janson dependency sum is
If , then . The first Janson inequality gives
For the reverse bound, the events that individual triangles are absent are decreasing, so Harris' inequality gives
where keeps the logarithmic estimate uniform. Thus the probability is .
If , the extended Janson bound gives
For a lower bound, fix a balanced bipartition of the vertices and require every edge inside either part to be absent. The resulting graph is bipartite, hence triangle-free, and this event has probability
Combining the bounds proves the second regime.

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