Bound $\|(A+B)-(A^{1/4}(1+A^{-1/2}BA^{-1/2})^{1/2}A^{1/4})^2\|$ in terms of commutator $\|AB-BA\|$ For positive definite matrices $A$ and $B$, can
$$
\|(A+B)-(A^{1/4}(1+A^{-1/2}BA^{-1/2})^{1/2}A^{1/4})^2\|
$$
be bounded in terms of $\|AB-BA\|$?
Note that if the matrices commute, then both norms are zero. 
 A: Let $[F,G] := FG-GF$.
$$
%(A^{1/4}C^{1/2}A^{1/4})^{2} = A^{1/4}C^{1/2}A^{1/2}C^{1/2}A^{1/4}\\
%= A^{3/4}C^{1}A^{1/4}+A^{1/4}[C^{1/2},A^{1/2}]A^{1/4}\\
%= AC+A^{1/4}[C^{1/2},A^{1/2}]A^{1/4}+A^{3/4}[C,A^{1/4}]
%$$
Easy algebra shows that 
$$
(A^{1/4}(1+A^{-1/2}BA^{-1/2})^{1/2}A^{1/4})^2= A+A^{1/2}BA^{-1/2}+A^{1/4}[C^{1/2},A^{1/2}]A^{1/4}+A^{3/4}[C,A^{1/4}]\\ = A+B+[A^{1/2},B]A^{-1/2}+A^{1/4}[C^{1/2},A^{1/2}]A^{1/4}+A^{3/4}[C,A^{1/4}].
$$
where $C:=(1+A^{-1/2}BA^{-1/2})$. Since
$$
[C,A] = CA-AC = A^{-1/2}[A,B]A^{-1/2}
$$
we get a a result if we can bound $[F^{r},G^{s}]$ in terms of $[F,G]$.
I don't have an elegant way to do this. One way is to note that there are polynomials $p$ and $q$ of degree $n$ such that $F^{r} = p(F)$ and $G^{s} = q(G)$, hence
$$
F^{r}G^{s} = p(F)q(G) - q(G)p(F) = \sum_{i,j=1}^{n}[F^{i},G^{j}]
$$
and we can focus on integer exponents. For those, repeatedly swapping $F$ and $G$ and collecting the commutators shows that
$$
\|[F^{i},G^{j}]\|\leq \|F\|^{n}\|G\|^{n}\|[F,G]\|.
$$
In summary, this gives us
$$
\|(A+B)-(A^{1/4}(1+A^{-1/2}BA^{-1/2})^{1/2}A^{1/4})^2\|\\ \leq \|[A,B]\| (  \|A\|^{n}\|B\|^{n}\|A^{-1}\|^{1/2}+\|A\|^{1/2}\|A\|^{n}\|C\|^{n}\|A^{-1}\|+\|A\|^{3/4}\|A\|^{n}\|C\|^{n}\|[A,B\|)
$$
with $\|C\|^{n}\leq \|A^{-1}\|^n2^{n}(\|A\|^{n}+\|B\|^{n})$
