TITLE: A bound on $e^{A+B}$ for noncommutative $A$ and $B$
QUESTION [8 upvotes]: It is folklore that when two complex $n\times n$ matrices $A$ and $B$ commute, the equality $e^{A+B}=e^Ae^B$ holds.  An elementary consequence of this is that for any submultiplicative norm, commuting matrices satisfy the inequality $$||e^{A+B}||\leq ||e^A||\cdot||e^B||.$$
Earlier this week, I was trying to prove a result (with $n=2$) that would have benefitted from the above inequality. However, my matrices did not commute. Testing a few thousand pairs of matrices in GNU Octave, not a single one failed to pass the test, which is reasonable evidence to suggest that the above inequality might hold generally or, if it fails, one might need to be clever in finding a counterexample.
I could not find a counterexample and did not get very far in proving it. 
While I have moved on a proved the result without the inequality, it's still bothering me. Could somebody shed some light on this?

REPLY [5 votes]: The inequality does not always hold. Counterexample:
$$
A=\pmatrix{0&-2\pi\\ 2\pi&0},
\ B=\pmatrix{0&-\pi\\ 4\pi&0},
\ A+B=\pmatrix{0&-3\pi\\ 6\pi&0}.
$$
Then $e^A=e^B=I$ and
$$
e^{A+B}=
\pmatrix{\cos\left(3\sqrt{2}\pi\right)&-\frac1{\sqrt{2}}\sin\left(3\sqrt{2}\pi\right)\\
\sqrt{2}\sin\left(3\sqrt{2}\pi\right)&\cos\left(3\sqrt{2}\pi\right)}.
$$
When the operator norm is used, numerically we have $\|e^{A+B}\|\approx1.2735>1=\|e^A\|\|e^B\|$.