If $A< C$ and $BI have symmetric positive definite matrices $A, B, C, D$ with
\begin{equation}
A \leq C \quad\text{and}\quad
B \leq D
\end{equation}
Is it true that $\|AB\|_F \leq \|CD\|_F$ where $\|\cdot\|_F$ denotes the Frobenius norm?
The closest question I could find was: Does Frobenius norm (not operator 2 norm) preserve the positive semidefinite order of matrices?. Based on this we have $\|A\|_F \leq \|C\|_F$ and $\|B\|_F \leq \|D\|_F$.
 A: No. Pick any non-diagonal matrix matrix $X$ such that $0<X<I$. Then $x_{ij}\ne0$ for some $i\ne j$. Let $C$ be the matrix obtained by modifying the $i$-th diagonal entry of the identity matrix to $t^2$, where $t>0$. The $(i,j)$-th element of $C^{1/2}XC^{-1/2}$ is then $tx_{ij}$. It follows that $\|C^{1/2}XC^{-1/2}\|_F>\|I\|_F$ when $t$ is sufficiently large.
Now let $A=C^{1/2}XC^{1/2}$ and $B=D=C^{-1}$. Then $A<C$ and $B\le D$ but
$$
\|AB\|_F
=\|(C^{1/2}XC^{1/2})C^{-1}\|_F
=\|C^{1/2}XC^{-1/2}\|_F
>\|I\|_F
=\|CD\|_F.
$$
For a concrete example, consider
$$
A=\pmatrix{32&2\\ 2&1},
\quad B=D=\pmatrix{\frac{1}{64}\\ &1}
\quad\text{and}\quad C=\pmatrix{64\\ &1}.
$$
Remark. However, in general we do have $\|\sqrt{A}\sqrt{B}\|_F\le\|\sqrt{C}\sqrt{D}\|_F$ because
$$
\begin{aligned}
\|\sqrt{A}\sqrt{B}\|_F^2
&=\operatorname{tr}(\sqrt{A}B\sqrt{A})\\
&\le\operatorname{tr}(\sqrt{A}D\sqrt{A})\\
&=\operatorname{tr}(\sqrt{D}A\sqrt{D})\\
&\le\operatorname{tr}(\sqrt{D}C\sqrt{D})\\
&=\|\sqrt{C}\sqrt{D}\|_F^2.
\end{aligned}
$$
Equivalently, the inequality $\|AB\|_F\le\|CD\|_F$ is true if we impose an additional condition that $A^2\le C^2$ and $B^2\le D^2$.
