Consider a symmetric positively definite (SPD) matrix $A^2$ and a sequence of SPD matrices $B^2_n$, such that $|tr{(A^2 - B^2_n)}| \leq \frac{C}{\sqrt{n}}$, where $C$ is some generic constant.
Is it possible to say anything about the upper bound $R_n$ on the quantity: $\Bigl|tr\bigl(AB_n + B_nA - 2(B_n A^2 B_n)^{1/2}\bigr)\Bigr| \leq R_n $,
or about a slightly modified version:
$\Bigl|tr\bigl(AB_n - (B_n A^2 B_n)^{1/2}\bigr)\Bigr| \leq R_n$,
e.g. that $R_n$ is of order $n^{-\alpha}$, $\alpha > 0$?
If there exist constants $R_n$ such that your inequality is satisfied, then the convergence $R_n\to0$ as $n\to\infty$ cannot follow from the inequality $|tr(A^2-B_n^2)|\le C/\sqrt n$.
Proof. Assume the contrary. Let $A$ and $B$ be SPD matrices. Define the sequence $B_n$ as follows: $B_n=\mu B$, where $\mu=\sqrt{tr A^2/(tr B^2)}$ (so that in fact $B_n$ is independent of $n$). Then $|tr(A^2-B_n^2)|=0\le 1/\sqrt n$. By our assumption, $|tr(AB_n+B_nA−2(B_nA^2B_n)^{1/2})|\le R_n\to0$ as $n\to\infty$ and hence $tr(AB_n+B_nA−2(B_nA^2B_n)^{1/2})=0$, because $B_n$ is independent of $n$. Thus, $tr(AB+BA−2(BA^2B)^{1/2})=\mu^{-1} tr(AB_n+B_nA−2(B_nA^2B_n)^{1/2})=0$. Thus, this equation holds for arbitrary SPD matrices $A$ and $B$, but this is obviously false; a contradiction.