Relationship between traces of products of certain matrices Let $A$ and $B$ be positive-definite Hermitian matrices.  
Does $\operatorname{Tr}[A^2B^{-1}]$ bound $\operatorname{Tr}[ABAB^{-2}]$, i.e., can one say $\operatorname{Tr}[A^2B^{-1}]\geq\operatorname{Tr}[ABAB^{-2}]$ or $\operatorname{Tr}[A^2B^{-1}]\leq\operatorname{Tr}[ABAB^{-2}]$ without knowing anything about the structures of $A$ and $B$ other than what's given above?
When $A$ and $B$ commute, the two traces are trivially equal, but I am stumped as to what happens when they do not.
 A: This thread is about a year old, but has no responses, so I've added one.  
claim:
$\tau_0= \text{trace}\big(A^2B^{-1}\big) \leq \text{trace}\big(ABAB^{-2}\big)=\tau_1$
for $\text{n x n}$ HPD $A,B$ with equality iff $BA=AB$ 
note:
$\tau_0=\Big\Vert B^\frac{-1}{2}A\Big \Vert_F^2\gt 0$
$\tau_1=\Big\Vert B^\frac{1}{2}A B^{-1}\Big \Vert_F^2\gt 0$
The proof proceeds (i) by applying Polar Decomposition, then (ii) Cauchy-Schwarz, and finally (iii) checking the equality conditions    
(i) applying Polar Decomposition to $B^{-1}A $
$AB^{-1}=PU^*\longleftarrow B^{-1}A = UP\longrightarrow A=BUP$ 
$\tau_0= \text{trace}\big(A^2B^{-1}\big)=\text{trace}\big(AB^{-1}A\big) = \text{trace}\big(\underbrace{B^{-1}A}A\big)=\text{trace}\big(UPA\big)=\text{trace}\big(UPBUP\big)$
$\tau_1 = \text{trace}\big(ABAB^{-2}\big)= \text{trace}\big(\underbrace{B^{-1}A}B\underbrace{AB^{-1}}\big)=\text{trace}\big(UPBPU^*\big)=\text{trace}\big(PBP\big)$ 
(ii) applying Cauchy-Schwarz
$\tau_0^2$
$=\Big\vert\text{trace}\big(UPBUP\big)\Big\vert^2$
$=\Big\vert\text{trace}\big(UPB^\frac{1}{2}B^\frac{1}{2}UP\big)\Big\vert^2$
$=\Big\vert\text{trace}\big((B^\frac{1}{2}PU^*)^*(B^\frac{1}{2}UP)\big)\Big\vert^2$
$\leq\text{trace}\big( UPBPU^*\big)\cdot \text{trace}\big(PU^*BUP\big)$
$=\text{trace}\big( PBP\big)\cdot \text{trace}\big(PU^*BUP\big)$
$=\tau_1\cdot \text{trace}\big(PU^*\underbrace{BUP}\big)$
$=\tau_1\cdot \text{trace}\big(\underbrace{PU^*}A\big)$
$=\tau_1\cdot \text{trace}\big(AB^{-1}A\big)$
$=\tau_1\cdot\tau_0$ 
thus
$\tau_0^2\leq \tau_1\cdot\tau_0 \longrightarrow \tau_0\leq \tau_1$ 
(iii) equality conditions
By Cauchy-Schwarz, the upper bound is met with equality iff
$B^\frac{1}{2}AB^{-1} =B^\frac{1}{2}\underbrace{PU^*}=\alpha\cdot B^\frac{1}{2}\underbrace{UP} = \alpha\cdot B^\frac{1}{2}B^{-1}A = \alpha\cdot B^\frac{-1}{2}A$ 
checking the trace:
$0\lt\text{trace}\big(B^\frac{-1}{4}AB^\frac{-1}{4}\big) =\text{trace}\big(B^\frac{1}{2}AB^{-1}\big)=\alpha\cdot\text{trace}\big( B^\frac{-1}{2}A\big)=\alpha\cdot\text{trace}\big( B^\frac{-1}{4}AB^\frac{-1}{4}\big)$
which implies $\alpha = 1$ 
Thus there is equality iff
$B^\frac{1}{2}AB^{-1} =B^\frac{-1}{2}A$
multiplying on the left by $B^\frac{1}{2}$ and on the right by $B$, this reads:
we have equality iff $BA =AB$ 
A: user8675309's answer is correct, but there is no need to use polar decomposition. With Frobenius inner product, Cauchy-Schwarz inequality implies that
$$
\operatorname{tr}(A^2B^{-1})^2
=\langle B^{-1/2}A,B^{1/2}AB^{-1}\rangle^2
\le\|B^{-1/2}A\|_F^2 \|B^{1/2}AB^{-1}\|_F^2
=\operatorname{tr}(A^2B^{-1})\operatorname{tr}(ABAB^{-2}).
$$
Since both $A$ and $B$ are positive definite, $\operatorname{tr}(A^2B^{-1})$ is positive. So, we conclude that $\operatorname{tr}(A^2B^{-1})\le\operatorname{tr}(ABAB^{-2})$. Equality holds if and only if one of $B^{-1/2}A$ or $B^{1/2}AB^{-1}$ is a scalar multiple of the other. By taking traces of both products, we see that this scalar multiple must be equal to $1$. Therefore equality holds iff $B^{-1/2}A=B^{1/2}AB^{-1}$, i.e. iff $AB=BA$.
A: Here's a much more succinct proof. Each matrix is $n\times n$.
$\text{trace}\big(A^2B^{-1}\big)=\Big\Vert B^\frac{-1}{2}A\Big \Vert_F^2 \leq \Big\Vert B^\frac{1}{2}A B^{-1}\Big \Vert_F^2=\text{trace}\big(ABAB^{-2}\big)$
Conjugation doesn't change commutativity and the Frobenius norm is unitarily invariant, thus we may instead prove
$\Big\Vert \Sigma^\frac{-1}{2}C\Big \Vert_F^2 \leq \Big\Vert \Sigma^\frac{1}{2}C \Sigma^{-1}\Big \Vert_F^2$
with $C:=Q^*AQ$ and
$Q^*BQ=\Sigma= \begin{bmatrix}\lambda_1 I_{r_1}& \mathbf 0&\cdots & \mathbf 0 \\ \mathbf 0 & \lambda_2 I_{r_2}&\cdots &\mathbf 0\\ \mathbf 0 & \mathbf 0&\ddots &\mathbf 0\\ \mathbf 0& \mathbf 0&\cdots & \lambda_m I_{r_m} \\\end{bmatrix}$
where $B$ has $m$ distinct eigenvalues.
For $i\neq k$ we have
$\big\vert c_{i,k}\big \vert^2 \cdot \big(\sigma_i^{-1}+ \sigma_k^{-1} \big) \leq \big\vert c_{i,k}\big \vert^2 \cdot \big(\sigma_i^{-2}\sigma_k+ \sigma_i \sigma_k^{-2} \big)$
by Muirhead's Inequality.  (The re-arrangement inequality also gives the result.). Summing over the bound:
$\Big\Vert \Sigma^\frac{-1}{2}C\Big \Vert_F^2$
$=\Big(\sum_{i=1}^n\sum_{k\gt i}^n \big\vert c_{i,k}\big \vert^2 \cdot \big(\sigma_i^{-1}+ \sigma_k^{-1} \big)\Big)+\sum_{i=1}^n \big\vert c_{i,i}\big \vert^2 \cdot \sigma_i^{-1}$
$\leq \Big(\sum_{i=1}^n\sum_{k\gt i}^n\big\vert c_{i,k}\big \vert^2 \cdot \big(\sigma_i^{-2}\sigma_k+ \sigma_i \sigma_k^{-2} \big)\Big)+\sum_{i=1}^n \big\vert c_{i,i}\big \vert^2 \cdot \sigma_i^{-1}$
$=\Big\Vert \Sigma^\frac{1}{2}C \Sigma^{-1}\Big \Vert_F^2$
re: equality conditions
if the upper bound is met with equality
$\sigma_i\neq \sigma_k\implies c_{i,k} = 0\implies  C = \begin{bmatrix}C_{r_1} & \mathbf 0&\cdots & \mathbf 0 \\ \mathbf 0 & C_{r_2}&\cdots &\mathbf 0\\ \mathbf 0 & \mathbf 0&\ddots &\mathbf 0\\ \mathbf 0& \mathbf 0&\cdots & C_{r_m}\\\end{bmatrix}$
Thus it is necessary (and obviously sufficient) that $C\Sigma = \Sigma C$ or equivalently, that $AB = BA$
