Assume we have two $n \times n$ real symmetric matrices $ A^2 $ and $B^2$, such that it holds for some $0\leq\rho<1$ $$ 0 < (1-\rho)B^2 \leq A^2 \leq (1+\rho)B^2, $$ where "$\leq$" means positive semidefinite ordering. Is it true, that for the operator norm we have $$ \| A^{-1} B\| \leq \frac{1}{\sqrt{1-\rho}} \ ? $$ As far as I understand, to obtain such bound we use only the inequality $(1-\rho)B^2\leq A^2$. Can we improve this bound knowing $A^2 \leq (1+\rho)B^2$?
Thanks!
EDIT: The answer is no. Conterexample: let $\rho = 0.99$, $$ A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}, \ \text{and} \ B = \begin{pmatrix} 20 & 0 \\ 0 & 1.1 \end{pmatrix}. $$ Then the conditions are fulfilled, but $$ \| A^{-1} B\| = \frac{1.022}{\sqrt{1-\rho}}. $$