Dirac's theorem states that if degree of each vertex of a graph $G=(V,E)$ is not less than $|V|/2$, then it has Hamiltonian cycle. It is less known, but still known and not so hard to prove (though I do not know the reference other than competition problem) that if all degrees are exactly $(|V|-1)/2$, the graph still have Hamiltonian cycle. So the question.

Let $G$ be a graph with $2n+1$ vertices and all degrees at least $n$ but without Hamiltonian cycle. What is minimal possible number of edges in $G$? There are two examples with $n(n+1)$ edges: $K_{n,n+1}$ and two $K_{n+1}$'s glued by a vertex. Maybe, this is least possible actually?

**UPDATE.** It looks that what actually is proved in the result to which I refer is that the graph on $2n+1$ vertices without Hamiltonian cycle and all degrees at least $n$ either contains $K_{n,n+1}$ or coincides with two $K_{n+1}$'s glued by a vertex. This answers the question, if anybody is interested I may leave a proof here, if not, just remove the question.