Graph theory undirected graph problem

Question

Let $G=(V,E)$ be an undirected graph , $|V|=n \ge 3$, n is Even. and $deg(v)\ge (n-1)/2$ $$$$ i need to show that G is an Hamilton graph... im here with about 5 other students... and no one has any clue... can someone please help me or show me a way to solve this problem? thank you!

Answer 1

The answer is simple

By Dirac (1952) :

A simple graph with n vertices ($n ≥ 3$) is Hamiltonian if every vertex has degree $\frac{n}{2}$ or greater. See https://en.wikipedia.org/wiki/Hamiltonian_path

Since n is even, then 2 does not divides $n-1$. Thus $$d(v)\geq \frac{n}{2}$$