Let $A, B\in M\_{n}(\mathbb{R})$ be symmetric positive definite matrices. It is easy to see $Tr(A^2+AB^2A)=Tr(A^2+BA^2B)$. Numerical experiments indicate $$Tr[(A^2+AB^2A)^{-1}]\ge Tr[(A^2+BA^2B)^{-1}],~~(1)$$ but it seems difficult to show it. 


Remark. When $n=2,3$, by direct computation, (1) is true. Here is an expriment done by matlab:


for s=1:1000



```
as=randn(4,4);
bs=randn(4,4);
ts=as*as';
rs=bs*bs';
ls=trace(inv(ts^2+ts*rs^2*ts)-inv(ts^2+rs*ts^2*rs))

```

end


{\bf Updated.} What about $A, B\in M\_{n}(\mathbb{C})$ be positive definite Hermitian matrices. 



A question on a trace inequality
Note first that $A^2+AB^2A=(A+iAB)(A-iBA)$. The reverse product is $(A-iBA)(A+iAB)=A^2+BA^2B-i(BA^2-A^2B)=X-iC$. Thus, the quantity on the left is $\operatorname{Tr} (X-iC)^{-1}$ and that on the right is $\operatorname{Tr} X^{-1}$. Moreover, the self-adjoint complex matrix $X-iC$ is positive definite (as the product of an invertible operator and its adjoint). Similarly, considering the factorization $A^2+AB^2A=(A-iAB)(A+iBA)$, we can write the quantity on the left as $\operatorname{Tr} (X+iC)^{-1}$. Symmetrizing, we see that it will suffice to show that $(X-iC)^{-1}+(X+iC)^{-1}\ge 2X^{-1}$ in the sense of quadratic forms (then the inequality for traces will hold too). We can multiply by $X^{1/2}$ from both sides to reduce it to $(I-iD)^{-1}+(I+iD)^{-1}\ge 2I$ where $D=X^{-1/2}CX^{-1/2}$ and both operators on the left are positive definite. Diagonalizing the self-adjoint operator $iD$, we see that the inequality reduces to $(1+p)^{-1}+(1-p)^{-1}\ge 2$ for $p\in(-1,1)$.


