# Is this trace inequality true?

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.

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There surely is some motivation behind all this? – Mariano Suárez-Alvarez Apr 25 '10 at 15:48
My purpose is to compare norms of two operators, it has been done. The updated one I created is more than what I want, I don't know how it occurs to me, but it looks interesting, so I post it. – Sunni Apr 25 '10 at 17:16