Show that in a graph $G$ if $\delta(G)\ge {n+1 \over 2}$ then every edge $e=(u,v)$ is a part of a Hamilton cycle.
Line of thinking: $G'=G/e$ has a Hamilton cycle by Ore's theorem ($d(u)+d(v)\ge n$ for all non-neighbors $u,v$). I would guess I need to somehow adjust the Hamilton cycle and then add back $e$ to it, but I don't see how. If I knew there was a Hamilton path from $u$ to $v$ in $G'$ obviously that would solve it.
HINT: Use Hamiltonian Closure. (I tend to use this since this is the only necessary and sufficient condition on Hamiltonian graphs known to me.)
A graph is Hamiltonian iff its closure is Hamiltonian.
Assume we want edge $uv$ in graph $G$ to be in a Hamiltonian cycle. So it suffices to prove that $u\ldots v$ is a Hamiltonian path. Add a new vertex $w$ and join to $u$ and $v$, forming a new graph $G_1$.
$\delta(G) \geq \frac{n+1}{2} $ implies $d(x) + d(y) \geq n+1 $ for vertices $x,y \in G$, hence $V(G)$ induces a clique in closure of $G_1$. After adding all edges in $G$, now since $d(w) + d(x) \geq n+1 $ for vertex $x \in G$, closure of $G_1$ is a clique, which is Hamiltonian. So $G_1$ is Hamiltonian, and there is a Hamiltonian path from $u$ to $v$ in $G$; hence your result.