In comparing the norm of two operators, I come across the following problem.


Let $S\in M\_{n}(\mathbb{R})$ be a symmetric matrix. $D\_1=diag(\alpha\_1,\cdots,\alpha\_n)$, $D\_2=diag(\beta\_1,\cdots,\beta\_n)$, with $\alpha\_1\ge\cdots\ge\alpha\_n\ge0, ~\beta\_1\ge\cdots\ge\beta\_n\ge 0$. Is it true that $Tr[(D\_1S^2D\_1D\_2+I)^{-1}]\le Tr[(SD\_1^2SD\_2+I)^{-1}]$?


**Updated** The above statment is not true as shown by Gerald Edgar. Now with the same notation, define $A=D\_2SD\_1^2SD\_2, B=D\_2D\_1S^2D\_1D\_2$ and
$$X\_{k+1}=\frac{1}{2}(X\_k+AX\_k^{-1}),~~ X\_0=I.$$
$$Y\_{k+1}=\frac{1}{2}(Y\_k+BY\_k^{-1}), ~~Y\_0=I.$$


Is it true $Tr X\_{k}\le Tr Y\_{k}$ for all $k\ge 1$?


When $k=1$, it is true.



Is this trace inequality true?
I just wrote down some random 2x2 example, and it had the wrong sense for the inequality.
A = [2,1;1,0], D1=D2=diag(2,1) .


