A tree has at least as many leaves as its maximum degree

1. First, a tree with $n\ge2$ vertices has a leaf: take a longest path $v_0v_1\dots v_m$ in the tree. The vertex $v_0$ has no neighbour outside the path and no neighbour $v_i$ with $i\ge2$ (that would create a cycle), so its only neighbour is $v_1$.

Now induct on $n$. For $n=2$ the tree is one edge. For $n>2$, remove a leaf and its edge: the remaining graph is still connected and acyclic, with $n-1$ vertices and, by induction, $n-2$ edges. So the original tree has $n-1$ edges.

2. Let $v$ be a vertex of degree $k\ge1$. If $k=1$, then the tree has at least one leaf (indeed at least $2$ when $n\ge2$), so the claim $L\ge k$ holds trivially. Assume $k\ge2$. Let $L$ be the number of leaves. Each leaf has degree $1$; $v$ has degree $k$; all other $n-L-1$ vertices are not leaves and have degree at least $2$. By the handshake lemma and part 1,
$$2(n-1)=\sum_u\deg u\ge L+k+2(n-L-1).$$
Simplifying the right side: $2n-L-2+k$. Therefore $2n-2\ge2n-L-2+k$, i.e. $L\ge k$.

Equality. It holds when all vertices other than $v$ and the leaves have degree exactly $2$, that is when the tree is a "spider": paths glued at $v$.

Lesson. Handshake lemma: relate the degrees to the number of edges.