The minimum edge cover of a tree is at least the maximum degree Let $T$ be a tree with maximum degree $\Delta(T)$, and let $\beta'(T)$ denote the size of the minimum edge cover of $T$. The question is to prove that $\beta'(T) \ge \Delta(T)$.
I started by proving that each tree has at least $\Delta(T)$ using induction on $n$. Then I tried using the fact that if we take the edge connected to each leaf then we will have minimum edge coverage equal to $\Delta(T)$ in case all the leaves are connected to the vertex of max degree, if not, then we need at least one more edge. Does it make sense?

Let $v$  be a vertex of a maximal degree $\Delta$ of the tree $T$. Let $N(v)$ be a set of neighbors of the vertex $v$. If any edge of $T$ is incident to at least two vertices $v,w$ of $N(v)$ then $u-v-w-u$ is a cycle, which cannot occur in a tree. So each edge cover need at least $|N(v)|=\Delta$ distinct vertices to cover $N(v)$.
