= Solution
First the <graph> is <connected>: every component has at least $\delta(G)+1>n/2$ vertices, so two components cannot fit. Take a <longest path> $v_1,\ldots,v_k$. Every neighbour of either endpoint lies on that path. In the index set $\{1,\ldots,k-1\}$ define
$$
I=\{i:v_1v_{i+1}\in E\},\qquad J=\{i:v_iv_k\in E\}.
$$
Their sizes sum to $d(v_1)+d(v_k)\ge n>k-1$, so they share an index $i$. The edges at that index give the cycle
$$
v_1,v_2,\ldots,v_i,v_k,v_{k-1},\ldots,v_{i+1},v_1,
$$
containing every vertex of the path. The minimum-<degree> assumption and $n\ge3$ ensure the resulting cycle is nondegenerate. If it omitted any <graph> vertex, <connectedness> would supply an edge from outside it to the cycle. Starting with that outside vertex and then following the cycle without closing it gives a path with $k+1$ vertices, contradicting maximality. Thus \b[the cycle is <Hamiltonian>]. This proves <Dirac's theorem> in the required case.
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