Induced-norm bound of composite operators. Let $A$ and $B$ be square real matrices with appropriate dimensions such that they do not commute. Generally, the following inequality does not hold.
$$\|ABx\|\leq\|B\|\|Ax\|$$
Is there any condition that makes it hold without assuming $AB=BA$?
 A: I'm going to interpret your question as follows:

Question: Under what conditions on $A$ and $B$ does $\|ABx\| \leq \|B\| \|Ax\|$ hold for all $x$?

I will work in the context of $n$-by-$n$ complex matrices, but everything works over the reals also.

Background: Recall a matrix $A$ is called Hermitian if $A^*=A$, where $A^*$ denotes the conjugate transpose. The spectrum of a Hermitian matrix real. A matrix is called positive semi-definite if $\langle Ax,x \rangle \geq 0$ for all $x$. A matrix is positive-semidefinite if and only if it is Hermitian and has nonnegative spectrum. The sum of two positive semi-definite matrices is positive semi-definite. The latter fact leads to a well-defined partial ordering of Hermitian matrices given by $A \leq B$ if and only if $B-A$ is positive semi-definite. Every positive semi-definite matrix $A$ has a unique positive semi-definite square root $A^{1/2}$. The square root operation is order preserving: $A \leq B$ implies $A^{1/2} \leq B^{1/2}$. Note however that the squaring operation is not order preserving. For any matrix $A$, the matrix $A^*A$ is positive semi-definite, hence has a square-root $|A| := (A^*A)^{1/2}$ which is sometimes referred to as the operator modulus.

The main fact I want to use is the following.

Lemma: Let $A$ and $B$ be matrices. Then, $\|Ax\| \leq \|Bx\|$ for all $x$ if and only if $|A|^2 \leq |B|^2$, with respect to the ordering on Hermitian matrices.


Proof: \begin{align*} 
&& \|Ax\| \leq \|Bx\| \text{ for all } x \\
& \Leftrightarrow & \langle Ax,Ax\rangle \leq \langle Bx,Bx\rangle \text{ for all } x \\
& \Leftrightarrow & \langle  (|B|^2 - |A|^2)x,x\rangle \geq 0 \text{ for all } x \\
& \Leftrightarrow & |B|^2 - |A|^2 \text{ is positive semi-definite} \\
& \Leftrightarrow & |A|^2 \leq |B|^2 \text{ with respect to the ordering on Hermitian matrices}
\end{align*}

A consequence of the above lemma is that, if $\|Ax\| \leq \|Bx\|$ for all $x$, then $|A| \leq |B|$ (by monotonicity of the square root). In particular, if $A$ and $B$ are positve-semidefinite, a necessary condition to get $\|Ax\| \leq \|Bx\|$ for all $x$ is to have $A \leq B$.
To get an answer to your question, we set $A$ equal to $AB$ and $B$ equal to $\|B\| A$ in the above lemma, which yields the following:

Corollary: Let $A$ and $B$ be matrices. Then $\|ABx\| \leq \|B\| \|Ax\|$ for all $x$ if and only if $|AB|^2 \leq \|B\|^2 |A|^2$ with respect to the ordering on Hermitian matrices. In particular, onc necessarily has $|AB| \leq \|B\| |A|$.

Note that the similar looking, but distinct, condition $|AB|^2 \leq \|A\|^2 |B|^2$ is actually always satisfied. This is because $|A|^2 \leq \|A\|^2$ always holds, where the right hand side is interpreted as a multiple of the identity matrix, and the operation $X \mapsto B^*XB$ is an order-preserving map of the set of positive-definite matrices.
