Suppose that $A$ and $B_1, B_2, \dots, B_n$ are square matrices with dimension $m$ and $x_1, \dots, x_n$ are vectors with dimension $m$. Let $B = [B_1, \dots, B_n] \in \mathbb{R}^{m \times nm}$. Does the following norm inequality hold?
\begin{align} \|A \sum_{i=1}^n B_i x_i\|_2^2 \leq \|B\|_F^2 \sum_{i=1}^n \|Ax_i\|_2^2 \end{align}
My attempt:
Suppose that $A$ and $B_i$ are commutative. Using Cauchy-Schwarz (not sure if the application of Cauchy-Schwarz here is correct)
$$\|A \sum_{i=1}^n B_i x_i\|_2^2= \|\sum_{i=1}^n B_i A x_i\|_2^2 \leq \sum_i \|B_i\|_2^2 \sum_i \|A x_i\|_2^2 \leq \|B\|_F^2 \sum_{i=1}^n \|Ax_i\|_2^2.$$
Is it possible to show this when $A$ and $B_i$ are not commutative?
No. E.g. when $n=1$, $$ A=\pmatrix{1&0\\ 0&0},\ B=B_1=\pmatrix{0&1\\ 0&0},\ x_1=\pmatrix{0\\ 1}, $$ we have $\left\|A\sum_iB_ix_i\right\|_2^2=1>0=\|B\|_F^2\sum_i\|Ax_i\|_2^2$.