Tree with no nodes of degree 2: prove that # leaves ># internal nodes using average degree and handshake lemma

Question

Im really struggling to formalise my thoughts on this one.

Basically I understand that if we would allow nodes with degree 2, then we could chain together infinitely many nodes to always produce more internal nodes than leafs.

Conversely, since we don't allow degree 2, every internal node contributes more than one child to the graph.

But this is where I get stuck. I don't know how to use the average degree to prove this inequality.

Any hints are welcome.

Thanks!

P.S Tree is not rooted

Answer 1

Hint Let $x$ be the number of leaves and $y$ the number of internal vertices. Then the total degree is at least

$$x+3y$$

By Handshaking Lemma (and tree formula), the total degree is exactly $$2(x+y-1)$$

Combine them.