Prove that in every tree on $n$ vertices there are at most ${n-2\over 3}$ vertices of degree more than three. I'm trying to show that: 

If $T$ is a tree on $n$ vertices, then there are at most ${n-2\over 3}$ vertices that have a degree greater than three. 

I've been unable to come up with anything so far that could be insightful or helpful. Any help? Thanks.
 A: Note  that  a tree on $n$ vertices has $n-1$ edges and therefore
$$\sum_{v\in V}\text{deg}(v)=2(n-1).$$
Let $a$ the number of vertices of degree greater than $3$, then
$$4a+(n-a)\leq \sum_{v\in V}\text{deg}(v)=2(n-1).$$
which implies 
$$a\leq \frac{n-2}{3}.$$
This approach can be easily generalized: if $a_d$ is the number of vertices of degree greater than $d\geq 1$ then
$$a_d\leq \frac{n-2}{d}.$$
A: PS: I have misread the question, my answer regards the case for vertices of degree greater or equal ($\geq$ ) $3$ and not $>3$:
First of all your formula is wrong, it should be $\frac{n-2}{2}$, e.g. take the 3-star (4 vertices all connected to the center).
To the proof:
Try to show it with induction: $n=1,2$ should be clear.
For bigger $n$ leave out a leave. This will yield a tree of lower degree. If you decrease the number of vertices of degree greater than 3, you now have a vertex of degree 2 (adjacent to the former leave). But you can now ignore this vertex also (by merging the to adjacent edges) and get a tree with 2 less vertices and the same amount of vertices of degree $\geq 3$. Apply now induction.
