Prove the number of vertices of degree 1 in an tree must be greater than or equal to the maximum degree in the tree. Prove the number of vertices of degree 1 in an tree must be greater than or equal to the maximum degree in the tree.
-My professor stated this as a fact in class, and said that we could find the accompanying proof online. I've tried to find it but was unsuccessful, I was wondering if anyone knows where I could find it or even how to do the proof itself, so that I could better understand why this fact is true.
 A: Let $T$ be a tree, and let $v$ be a vertex of $T$ of maximal degree; say $\deg v=n$. Let the $n$ vertices adjacent to $v$ be $v_1,\ldots,v_n$. Remove $v$ from $T$; $T-v$ has $n$ components, say $T_1,\ldots,T_n$, where $T_k$ contains the vertex $v_k$ for $k=1,\ldots,n$. Each of those components is a tree, so it has at least two vertices of degree $1$; that means that $T_k$ has a vertex $u_k\ne v_k$ whose degree in $T_k$ is $1$. But $\deg_{T_k}w=\deg_Tw$ for every vertex of $T_k$ except $v_k$, so $\deg_Tu_k=1$. Thus, the vertices $u_1,\ldots,u_n$ are $n$ vertices of $T$ of degree $1$.
A: Take a spanning tree for the tree. Contract all edges that join two vertices of degree > 1 to a single point, resulting in one 'main' vertex and a whole lot of edges sticking out from it to all the verts of degree 1. 
That main vertex must have degree at least as large* as the greatest degree of any node in the original tree, but in the reduced tree, its degree is exactly the sum of the degrees of the degree-1 vertices. 


*

*proof of this: consider an edge-by-edge contraction. If we contract an edge PQ, the degrees at P and Q, $d_P$ and $d_Q$, and both at least 2, but the degree of the merged vertex is $d_P + d_Q - 2$ (the "2" being for the lost edge). Since $d_Q - 2 \ge 0$, this is at least as large as $d_P$; a similar argument holds for $d_Q$. 

