trace inequalities: linear algebra

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If S is any $n \times n$ real, symmetric, invertible matrix and D is any $n \times n$ diagonal matrix such that $0\prec D \prec I$ then does there exist a constant $\gamma$ such that:

$trace(S^{-1}DS^2DS^{-1})\leq \gamma \cdot n$

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If you are asking whether there exists a constant $\gamma$, that is independent of $n,S$ and $D$, such that $\operatorname{trace}(S^{-1}DS^2DS^{-1})\le\gamma\,n$, the answer is no. Let $x>0$ and $$ D=\frac14\pmatrix{1\\ &2},\ U=\frac1{\sqrt{2}}\pmatrix{1&1\\ 1&-1},\ S=U\pmatrix{1\\ &x}U^T. $$ Then $S^{-1}DS=\frac14U\pmatrix{\frac32&-\frac x2\\ -\frac1{2x}&\frac32}U^T$ and hence $$ \operatorname{trace}(S^{-1}DS^2DS^{-1}) =\left\|\frac14\left(\frac32,\,-\frac1{2x},\,-\frac x2,\,\frac32\right)\right\|^2, $$ which is unbounded as $x\to0$ or $x\to\infty$.