I will use $\|\|_*$ to denote the nuclear norm (sum of singular values) and $\|\|_2$ to denote the operator norm / matrix 2-norm (largest singular value).

Consider two positive definite $n \times n$ matrices $A$ and $B$ such that $\text{trace}(A)=\text{trace}(B)\equiv k$. I am interested in finding the largest constant $c \geq 0$ such that the following inequality holds:

$$ \|\sqrt{B}^{-1} A \sqrt{B}^{-1}\|_2 \geq c \|A - B\|_*.$$

Unable to obtain a proof, I have done some numerical investigation and I have noticed that it's easy to get violations of the inequality if the trace of the matrices $k$ is chosen to be much larger than $n$ (with fixed $c$). However, as the trace decreases, so do the violations, and for smaller values of $k$ I was not able to find numerical counterexamples unless the value of $c$ gets very large.

I am curious about whether, given $n$ and $k$, there is a value of $c$ such that the inequality always holds. Or perhaps whether the inequality always holds for some $n$-dependent $c$ when one simply sets $k=1$.

I would appreciate any thoughts and ideas.