prove that graph with girth=5 has at least $k^2+1$ vertices

Question

given a graph G with a girth of 5, prove that for $delta(G)>=k$ (minimum degree), G has a least $k^2+1$ vertices.

Can anyone provide a hint or an approach for this?

Thanks

Answer 1

Pick

• a vertex,
• its neighbours,
• and their neighbours apart from the original vertex