Problem
$G$ is a simple graph with $n$ vertices. Its minimum degree $\delta(G) \geqslant \frac{n+q}{2}$.
The following set of edges $E_0=\{e_1,...,e_q\} \subset E(G)$ has $2q$ distinct vertices.
Prove that there exists a Hamilton cycle of G which contains all the edges of $E_0$.
My Thoughts
$\delta \geqslant \frac{n}{2}$, so $G$ is Hamiltonian.
Proof by contradiction:
Suppose the Hamilton cycle $H$ of $G$ contains the most number of edges from $E_0$, but $\exists e \in E_0$ s.t. $e\notin H$. Denote $H$ as $v_1v_2...v_n$, $e$ as $v_1v_k$.
$v_n$ cannot be connected to $v_{k-1}$. $v_2$ cannot be connected to $v_{k+1}$. Otherwise there will be a Hamilton cycle $H'$ that contains more edges from $E_0$ than $H$.
And I'm stuck here.