Set and take . Its expected number of edges is asymptotic to , and a Chernoff bound makes with positive probability. The expected number of cycles of length below is at most
which is far below ; Markov's inequality shows that, simultaneously with positive probability, there are fewer than such cycles.
Choose such a graph and delete one edge from every cycle of length below . The resulting graph has girth at least and more than edges, hence average degree greater than . Part (a) supplies a subgraph of minimum degree at least . It uses at most vertices and cannot acquire any shorter cycle.

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