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**zhupolongjoe** Prove that if G is a graph of order n such that delta(G)>=(n-1)/2, then lambda(G)=delta(G).

Well, because delta(G)>=(n-1)/2, we know G is connected from a previous result. Also, this bound was proven to be sharp. We also know by a theorem that delta (G)<=lambda(G) for every graph G. So if we could just show that it is not the case that delta (G) < lambda(G), we would be done.

This is where I am stumped. I am pretty sure it involves the inequality with delta(G), but I don't know what to do with it....