In a Tree, show that the largest degree of a node <= number of nodes of degree 1 Let $T$ be a tree in which the largest degree of a node equals to $t$. Let $n_1$ denote the number of nodes of degree $1$ in $G$. Prove that $n_1 ≥ t$
I understand why this works but I am not sure how to prove it mathematically. It makes sense, because vertices of degree one are those at the end of each leaf (let their number be n) and/or the vertex in the beginning of the tree that doesn't branch into more than one edge. And the vertex with highest degree is gonna have at max n edges connected to it. Am I making sense? any help in the formal proof?
 A: Your inquality is certainly true for $n \leq 2$. For bigger trees, consider the following lemma:
$$\#\text{leaves} = 2 + \sum_{\deg(v)\; \geq\; 2}\Big(\deg(v)-2\Big).\tag{$\spadesuit$}$$
As all the entries in the sum above are non-negative, so we obtain
$$\#\text{leaves} \geq 2 + \max_{\deg(v)\; \geq\; 2}\Big(\deg(v)-2\Big) = \max_{v \in V} \deg(v)$$
which is exactly what we need.
Proof of the $\spadesuit$ equality:
We know that the sum of all degrees is twice the number of edges and the number of edges in any tree is the number of vertices minus one, so 
$$\sum_{v\in V}\deg(v) = 2m = 2(n-1).$$
Naturally $\sum_{v \in V}1 = n \cdot 1 = n$, hence
$$\sum_{v \in V}\Big(\deg(v)-2\Big)= -2.$$
Splitting $V$ into leaves and non-leaves we get
$$\sum_{\deg(v) = 1}\Big(\deg(v) - 2\Big) + \sum_{\deg(v) \geq 2}\Big(\deg(v) - 2\Big) + 2 = 0,$$
that is,
$$\sum_{\deg(v) = 1}\Big(-1\Big) + \sum_{\deg(v) \geq 2}\Big(\deg(v) - 2\Big) + 2 = 0,$$
simplifying
$$\sum_{\deg(v) \geq 2}\Big(\deg(v) - 2\Big) + 2 = \sum_{\deg(v) = 1}\Big(1\Big) = \#\text{leaves}.$$
$$\\\tag*{$\square$}$$
I hope this helps $\ddot\smile$
A: Consider the tree to be rooted with the root being a vertex of maximum degree. Each edge from the root must lead to a different leaf (perhaps to more than one leaf).
A: Let $v$ have degree $t$ and neighbours $v_1,\ldots, v_t$. 
For each vertex $u\ne v$ let $\ell(u)$ denote the length of the (in a tree: uniqe) shortest path from $u$ to $v$, and let $\phi(u)\in\{v_1,\ldots,v_t\}$ be the last node before $v$ in this path.
Let $S_i=\{\,u\ne v\mid \phi(u)=v_i\,\}$ for $1\le i\le t$.
Then the $S_i$ are pairwise disjoint and nonempty (because $v_i\in S_i$).
For each $i$, pick $u_i\in S_i$ with $\ell(u_i)=\max\{\,\ell(u)\mid u\in S_i\,\}$.
Show that $u_i$ is a leaf.
