Worst case bisection of binary tree
Nobody said that $3/4$ was the optimal number. In fact, it seems like $2/3$ might be achievable. Starting at the root, consider a walk which always chooses the child with the larger subtree. Consider the first time that you reach a node whose subtree contains less than $2/3$ of the nodes. Its parent's subtree contains more than $2/3$ of the nodes, so the subtree rooted at its larger child contains at least $1/3$ of the nodes. So we have found a node whose subtree contains between $1/3$ and $2/3$ of the nodes.

Considering the tree with one root and two children, we see that we can't in general obtain a better balanced partition.