Let G = (V, E) be a tree. Prove that at most one vertex can have degree at least |V |/2 + 1.
I tried to solve this by using a proof by contradiction. I assumed that at least two vertices can have a degree of |V |/2 + 1. But It didn't help that much. Any ideas?
Let $v_1,v_2$ be two vertices of the tree. Then there is a unique path $v_1,...,v_2$. Remove any of the edges of this path. We get two connected components, each containing one of $v_1$ and $v_2$. At least one of the two connected components contains no more than $|V|/2$ vertices. The one $v_1$ or $v_2$ in that component will have to have degree no more than $|V|/2$.