I am working to prove a good upper bound on a matrix Frobenius norm. In what follows, I have a positive definite matrices $A \in \mathcal{S}^n_{++}$, and two symmetric matrices $r$ and $R$ where I am working under the assumption that the size of $r$ is controlled by $R$, that is; $$ \| r \|_F \leq \beta \| R \|_F,$$ where $\beta \in (0,1)$.
Now, in my work I have a term of the form $$ \| A^{-1/2} r A^{1/2} \|_F$$ where $A^{1/2}$ denotes the symmetric square root of $A$. However, the only assumption I am making on $r$ is that it is a symmetric matrix, and I do not wish to make any further simplifying assumptions. Thus, is there a way to show: $$ \| A^{-1/2} r A^{1/2} \|_F \leq \beta \| A^{-1/2} R A^{1/2} \|_F?$$
I think we can use the fact that $\| \cdot \|_F$ is submultiplicative to arrive at: \begin{align*} \| A^{-1/2} r A^{1/2} \|_F &\leq \| A^{-1/2} \|_F \|r\|_F \|A^{1/2} \|_F \\ &= \kappa_F (A^{1/2}) \| r \|_F \\ &\leq \beta \kappa_F (A^{1/2}) \| R\|_F \\ &= \beta \| A^{-1/2} \|_F \| R \|_F \| A^{1/2} \|_F <\| A^{-1/2} \|_F \| R \|_F \| A^{1/2} \|_F, \end{align*} where $\kappa_F (\cdot)$ denote the Frobenius condition number of a matrix. However, I am not sure if this gives precisely the conclusion that I need.
The inequality does not hold.
For instance, let $$ r=\begin{bmatrix} 0&1/2\\1/2&0\end{bmatrix},\qquad R=\begin{bmatrix} 1/2&1/3\\1/3&1/2\end{bmatrix},\qquad A=\begin{bmatrix} 1&0\\0&4\end{bmatrix}. $$ Then $$ \|r\|_F=\frac{\sqrt2}2<\frac{\sqrt{13}}{\sqrt{18}}=\|R\|_F, $$ while $$ \|A^{-1/2}rA^{1/2}\|_F=\frac{\sqrt{17}}4>\frac{\sqrt{35}}6=\|A^{-1/2}RA^{1/2}\|_F $$