Question
Let $\gamma$ be the length of longest simple path in graph $G$. If $G$ has only one vertex with minimum degree $\delta$, prove $\delta \le \gamma - 2 $
It's trivial to prove that $\delta \le \gamma$ for all graphs and when $\delta=\gamma$ then $G$ must have at least two vertices with minimum degree.
However I have trouble to prove that the case $\gamma - 1 = \delta$ is not possible.
Source Problem A51 from Graph Theory A Problem Oriented Approach
Edit 1 (adding more context)
The two previous problems in books (A49 and A50) are about proving that endpoints of longest path have degree less than $\gamma$ and proving that a simple path with length $\delta$ exists.
The first problem can be solved by showing that endpoints of longest simple path can only be adjacent to vertices of the path So at most their degrees will be equal to $\gamma$.
Using this result we can prove the theorem for the next problem that $\delta \le \text{degree of end point vertices}\le \gamma$.