= Solution
Let $v_1\cdots v_k$ be a longest <path>. Every neighbour of either endpoint lies on it. Among the $k-1$ indices $i=1,\ldots,k-1$, mark those with $v_1v_{i+1}$ an edge and those with $v_iv_k$ an edge. Their total number is
$$
\deg v_1+\deg v_k\geq n>k-1,
$$
so some index has both marks. The edges at that index close the path into the <cycle graph>
$$
v_1v_2\cdots v_iv_kv_{k-1}\cdots v_{i+1}v_1.
$$
The minimum-degree hypothesis also forces the graph to be a <connected graph>: each component would otherwise have at least $n/2+1$ vertices. If $k<n$, an edge from this cycle to an outside vertex would give a longer path by opening the cycle at its incident vertex. Hence $k=n$, proving the <Dirac theorem> and \b[Hamiltonicity].
The bound is sharp. For $n=2m$, take two disjoint copies of $K_m$, with minimum degree $m-1$. For $n=2m+1$, take two copies of $K_{m+1}$ sharing exactly one vertex. The minimum degree is $m$, but the shared vertex is a <cut vertex>, which no Hamiltonian graph can have. Both constructions have
$$
\boxed{\delta=\lceil n/2\rceil-1.}
$$
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