TITLE: How many vertices of degree 3 or more can a tree have at most?
QUESTION [5 upvotes]: It is known that a tree $T=(V,E)$ has at least $\Delta$ leaves, where $\Delta$ is the maximum degree of $T$. But how many vertices of specific degree at least $k$ can a tree have at most? I'm particularly interested in $k=3$, but also in general. Is there an easy way to see or count this? 
Let $s$ be the number of vertices of degree 3 or more in any tree. I think it's easy to see that $s < |V|$, but this is probably not tight.

REPLY [5 votes]: If we denote the number of vertices as $n$, from the hand shaking lemma and the fact that the number of edges in a tree is $n-1$, we have
$$\sum_{v \in V} d_n = 2 \vert E \vert = 2(n-1)$$
If $s_k$ vertices have degree at-least $k$, then we have
$$2(n-1) = \sum_{v \in V} d_n \geq s_k \times k +(n-s_k)$$
Hence, we get that
$$s_k(k-1) \leq n-2 \implies s_k \leq \dfrac{n-2}{k-1} \implies s_k \leq \left\lfloor\dfrac{n-2}{k-1} \right\rfloor$$
This bound is tight, i.e., you can construct a tree with exactly $\left\lfloor\dfrac{n-2}{k-1} \right\rfloor$ vertices of degree $k$.
Hence, in the case of $k=3$, we have $s_3 \leq \left\lfloor\dfrac{n}2 \right\rfloor - 1$.
I hope I understood your question correctly.