The proof that a graph G with n vertices, where the sum of the degrees of any two vertices is at least n, has a Hamiltonian path is based on Ore's Theorem. However, this doesn't necessarily hold for a multigraph since Ore's theorem applies to simple graphs without loops or multiple edges.
The question asks to prove that if G is a graph with n vertices and the sum of the degrees of any two vertices is at least n, then G has a Hamiltonian path. To prove this, one can use Ore's Theorem, which states that if G is a simple graph with n vertices (where n ≥ 3) and the sum of degrees of any two non-adjacent vertices is at least n, then G is Hamiltonian.
Since in our graph the condition is stronger (it applies to any two vertices, not just non-adjacent), G is certainly Hamiltonian. The same cannot be necessarily said for a multigraph, which permits multiple edges between two vertices and loops, because Ore's Theorem applies to simple graphs.
Therefore, given the hypothesis that the sum of degrees of any two vertices is at least n, one can assert that G possesses a Hamiltonian path due to the application of Ore's Theorem. This proof, however, does not necessarily extend to multigraphs, where additional considerations about the edges may affect the graph's Hamiltonicity.