Cauchy-Schwarz for sums of products of matrices The usual Cauchy-Schwarz inequality states that, for real sequences $a_i,b_i$
$$
\Big|\sum_{i=1}^n a_i b_i \Big|
\leq \Big(\sum_{i=1}^n a_i^2\Big)^{1/2}\Big(\sum_{i=1}^n b_i^2\Big)^{1/2}.
$$
My question is whether the same holds for matrices.  More precisely, let $A_i,B_i$ be sequences of $m\times m$ matrices.  Does it hold that
$$
\Big\|\sum_{i=1}^n A_i B_i \Big\|
\leq 
\Big\| \sum_{i=1}^n A_i^*A_i\Big\|^{1/2}
\Big\| \sum_{i=1}^n B_i^*B_i\Big\|^{1/2},
$$
where $\|\cdot\|$ is the operator norm?
 A: The Cauchy–Bunyakovsky–Schwarz inequality does not generalise as proposed above.
For $m=2=n$ consider the concrete choices
$$A_1=\begin{pmatrix}1 &0\\1&0\end{pmatrix},\;
B_1=\begin{pmatrix}1 &0\\0&0\end{pmatrix},\;
A_2=\begin{pmatrix}0 &1\\0&1\end{pmatrix},\;
B_2=\begin{pmatrix}0 &0\\0&1\end{pmatrix}.$$
A: For posterity's sake, I wanted to add a correct generalization of the inequality.
\begin{align*}
\Big\|\sum_{i=1}^N A_iB_i\Big\|
&= \sup_{x,y} \sum_{i=1}^N \langle x,A_iB_i y\rangle \\
&= \sup_{x,y} \sum_{i=1}^N \langle A_i^*x,B_i y\rangle \\
&\leq \sup_{x,y} \sum_{i=1}^N \|A_i^* x\|\|B_i y\| \\
&\leq \sup_{x,y} \Big(\sum_{i=1}^N \|A_i^* x\|^2\Big)^{1/2}
\Big(\sum_{i=1}^N \|B_i x\|^2\Big)^{1/2} \\
&= \sup_{x,y} \Big(\sum_{i=1}^N \langle x, A_iA_i^* x\rangle\Big)^{1/2}
\Big(\sum_{i=1}^N \langle x, B_i^*B_i x\rangle\Big)^{1/2} \\
&= \Big\|\sum_{i=1}^N A_iA_i^*\Big\|^{1/2}
\Big\|\sum_{i=1}^N B_i^*B_i\Big\|^{1/2}.
\end{align*}
Note that the only difference is that one must replace $A_i^*A_i$ by
$A_iA_i^*$.   This corrected inequality is compatible with Hanno's example.
