I'm looking at a proof of the above statement. It goes as follows:
For any graph $G = (V, E)$, write $l$ for the number of leaves. From the Handshaking lemma:
$2|E| = \sum_{v \in V} d(v)$
$=l+\sum_{v\in V, d(v)\geq 2} d(v)$
$\geq l + \triangle(G)+2(|V|-l-1)$
where $\triangle(G)$ is the maximum degree of the graph.
I don't understand where the last inequality comes from.
There are $|V|-l$ vertices that are not leaves. One of them, which we denote $u$, has to have degree $\Delta(G)$ while the other $|V|-l-1$ has degree at least 2. Hence $\sum_{v\in V,d(v)\geq 2} d(v) = \Delta(G) + \sum_{v \in V\setminus \{u\}, d(v)\geq 2} d(v) \geq \Delta(G) + 2(|V|-l-1)$