Vertices in tree graphs

Question

Prove that any tree with n nodes has at least n/2 nodes with degree 1 or 2.

I was able to establish that there are at least 2 nodes of degree 1 in any tree. This is the most I've progressed so far however.

Answer 1

In a tree with $n$ nodes, there are a total of $n - 1$ edges, so the sum of degrees is $2 n - 2$ and the average degree is $2 - \frac{2}{n}$. Since the tree is connected, each node has degree at least $1$. If $\frac{n}{2} + 1$ nodes or more had degree 3 or more, then the total degree would be at least $$3 \left(\frac{n}{2} + 1 \right) + 1 \left(\frac{n}{2} - 1 \right) = 2 n + 2$$

Answer 2

Let's say there are $\alpha$ vertices of degree 1 or 2. Suppose (by contradiction) that $\alpha<\frac{n}{2}$.

Then, \begin{align*}\sum_{v\in V}d(v)&=\sum_{v\in V, 1\leq d(v)\leq 2}d(v)+\sum_{v\in V,d(v)\geq 3}d(v)\\&\geq \sum_{v\in V, 1\leq d(v)\leq 2}1+\sum_{v\in V,d(v)\geq 3}3\\&=\alpha+(n-\alpha)\cdot 3\\&=3n-2\alpha\end{align*}

Can you find a contradiction and finish the proof?