Is it true that every graph with $n$ vertices in which $\delta(G)\geq\frac{n}{2}-1$ has Hamiltionian path?

Question

Is it true that every graph with $n$ vertices in which $\delta(G)\geq\frac{n}{2}-1$ has Hamiltionian path? $\delta(G)$ is the minimum degree of $G$.

Answer 1

A laconic answer.

Y

An explanation of the laconic answer.

“Y” means No, that is a counterexample is a star $K_{1,3}$.

A bonus answer. On the other hand, if $n\ge 3$ and $\delta(G)\ge \frac n2$ then $G$ has a Hamiltonian cycle (see, for instance, Corollary at the last page of slides of the lecture “Round trips” in Algorithmic Graph Theory by Joachim Spoerhase and Alexander Wolff). This also allows us to prove if $n\ge 2$ and $\delta(G)\ge \frac n2-\frac 12$ then $G$ has a Hamiltonian path. Indeed, let $G^*$ be $G$ with an additional vertex $v^*$ adjacent to every vertex of $G$. Then $$\delta(G^*)=\delta(G)+1\ge \frac n2+\frac 12=\frac{n+1}{2}.$$ Thus $G^*$ has a Hamiltonian cycle. When we remove the vertex $v^*$ from the cycle we obtain a Hamiltonian path for $G$.