proof non-empty AVL tree
If the tree hasn't any single-children the hypothesis trivially holds.
Given a single-child vertex $v$, we can show that the parent of this vertex $v$ cannot be a single-child.
Let $p(v)$ be the parent of $v$.
If $p(v)$ is the root then $p(v)$ isn't a single-child per definition.
If $p(v)$ is not the root: Assume that $p(v)$ is a single-child then $p(p(v))$ has only one child. This implies that the difference in height of the subtrees of $p(p(v))$ must be at least $2$ which conflicts with the AVL-criterium and therefore implies that $p(v)$ is not a single-child.
Using this and the fact that for every single-child there is a unique parent, we can conclude that $2\cdot numOnly \leq n$.