Minimum number of nodes in AVL tree
Prove a lower bound of 7 vertices:

Since the tree is of height 3, it must have a path of length 3 from the root to a leaf, so we already have to have 4 vertices; $r-v_1-v_2-v_3$.  Now r has to have balance factor in $\{1,0,-1\}$ (i.e. $|\text{BF}(r)|\leq1$) so it must have a path of length at least two on it's other subtree, so we have $r-v_4-v_5$ (at least two more vertices).

For the same reasons (BF) $v_1$ has to have at least one child on it's other subtree, so we have $v_1-v_6$ (at least one more vertex).

So we have proved that an AVL tree of height 3 has to have at least 7 vertices, your example shows that there is an AVL tree with 7 vertices, so the minimum is 7.