How do I show the following result for two commuting bounded idempotent linear operators on a normed space? I'm trying an old qualifier problem stating that for two commuting idempotent bounded linear operators $A$ and $B$ on a normed linear space, either $A=B$ or the operator norm of $A-B$ is at least one.
I'm at a complete loss as to how to approach this. Playing round with norms doesn't seem to get me anywhere and I don't see what other theorem to employ. 

From $A^2=A, B^2=B$ and $AB=BA$ we get
$(A-B)^3=A-B$ . Let $P:=(A-B)^2$. Then $P^2=P$.
Case 1: $P=0$, hence $A+B=2AB$. Multiplying with $A$ gives: $A+AB=2AB$, thus $A=AB$. We have the same result for $B$: $B=AB$. Hence $A=B$
Case 2: $P \ne 0$. Then: $||P||=||P^2|| \le ||P||^2$ and so $||P|| \ge 1$.
Consequence: $||(A-B)^2|| \ge 1. $ This gives
$1 \le ||(A-B)^2|| \le ||A-B||^2$. Therefore $||A-B|| \ge 1$
