Inequality for the operator norm of a product of matrices

I am working with a product of $$n\times n$$ matrices $$A_1,\ldots,A_k$$. Under which conditions can I assume that

$$\|A_1\cdots A_k\|_\infty \leq \|A_1\cdots \hat{A_i}\cdots A_k\|_\infty \|A_i\|_\infty,$$

where $$\|\cdot\|_\infty$$ denotes the operator norm, and $$\hat{A_i}$$ denotes the omission of $$A_i$$.

E.g. if all products of subsets of $$\{A_1,\ldots,A_k\}$$ are normal, then the above inequality should follow from the submultiplicativity of the operator norm and the fact that $$AB$$ and $$BA$$ have the same singular values for normal $$A$$ and $$B$$. However, this condition seems rather artifical and I am wondering if something more powerful holds. What if all $$A_i$$ are normal, self-adjoint, unitary or orthogonal projectors?

For a simple counterexample, suppose $$A_1$$, $$A_2$$, $$A_3$$ are the orthogonal projections $$A_1 = \pmatrix{1 & 0\cr 0 & 0\cr},\ A_2 = \pmatrix{1/2 & 1/2\cr 1/2 & 1/2\cr},\ A_3 = \pmatrix{0 & 0\cr 0 & 1\cr}$$ Then $$\|A_1 A_3\| = 0$$ but $$\|A_1 A_2 A_3\| > 0$$.
That covers all your cases except "unitary". Of course if the $$A_i$$ are all unitary, both sides of your inequality are $$1$$.
• A famous example from quantum mechanics! $A_1$ is vertical polarization, $A_2$ is diagonal, and $A_3$ is horizontal. $A_1$ followed by $A_3$ blocks everything, but if you interpose $A_2$ then something gets through. – Nik Weaver Jan 21 at 19:01