Ternary decision tree for the lower bound of comparison sort with 3-way comparisons
The proof would be exactly the same, using a ternary tree (where each internal node has three children) instead of a binary tree. A ternary tree of depth $d$ has at most $O(3^d)$ notes, so a sorting tree would have to satisfy $n! = O(3^d)$, which implies that $d = \Omega(n\log n)$.