Let $n$ be the number of the vertices of a graph $G$. If the degree of each vertex of $G$ is at least $\frac{n+1}{2}$, prove that for an edge $e$ of $G$ there is a Hamiltonian cycle containing $e$.
I know that the theorem which says that if for every two vertices $x,y$ of a graph, $d(x)+d(y)\ge n$ then there is a Hamiltonian path. I tried to use the idea of the proof of this theorem, using maximal paths, but I could not conclude this assertion.
If $G$ is an $n$-vertex graph with minimum degree $\frac{n+1}{2}$, then take any edge $uv$ and subdivide it: create a new vertex $x$, delete edge $uv$, and add edges $ux, xv$. The result is an $(n+1)$-vertex graph where $x$ has degree $2$ and every other vertex still has degree at least $\frac{n+1}{2}$. It suffices to prove that the new graph is Hamiltonian: if so, the Hamiltonian cycle must use edges $ux$ and $xv$ when it visits $x$, and so we can turn it into a Hamiltonian cycle in $G$ that contains edge $uv$.
A theorem of Chvátal gives what is essentially the best possible degree condition for Hamiltonian cycles. It is usually stated for $n$-vertex graphs, but I will state it for a $p$-vertex graph, because we are about to apply it to an $(n+1)$-vertex graph:
In our case, $\deg(v_1) = 2$ (because $v_1 = x$) and $\frac {n+1}2 \le \deg(v_2) \le \dots \le \deg(v_{n+1})$. The condition $\deg(v_k) \le k$ just barely does not hold when $k=1$; from there, it cannot hold for any $k < \frac{n+1}{2}$, so Chvátal's condition is always satisfied. (That is, $P(k) \implies Q(k)$ is always true because $P(k)$ is never true.)
So the $(n+1)$-vertex graph has a Hamiltonian cycle, and therefore the original $n$-vertex graph has a Hamiltonian cycle containing any edge $uv$ we choose.