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Let ${\bf A} \in \mathbb{R}^{n \times n}$ be symmetric positive definite.

Can one prove the following inequality for some positive constant $b$? \begin{align} {\rm Tr} ( {\bf A} ({\bf I}+ b {\bf A})^{-1}) \le \frac{n {\rm Tr}({\bf A})}{n +b {\rm Tr}({\bf A}) } \end{align}

Edit: Based on one of the answers below the question boils down to showing the following inequality \begin{align} \sum_{i=1}^n \frac{d_i}{1+b d_i } \le \frac{ n \sum_{i=1}^n d_i}{\sum_{i=1}^n (1+b d_i) } \end{align} where $d_i \ge 0$.

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If $A=aI,\ a>0$ then the inequality holds

If $A=SDS^T$ where $S$ is orthogonal and $D$ is diagonal not multiple of identity, then we have a claim $$ f(b):=\sum_i \frac{d_{ii}}{1+bd_{ii}} -n \frac{\sum_i d_{ii}}{\sum_i (1+bd_{ii})} \leq 0$$ where $D={\rm diag}\ (d_{11},\cdots, d_{nn})$

Note that $f(0)=0$ and $$f'(b=0)=-\sum_i d_{ii}^2 + \frac{1}{n}(\sum_i d_{ii})^2<0$$

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    $\begingroup$ Just wanted to clarify. How you concluded that $f(b) \le 0$ for all $b \ge 0$. How did you use $f(0)=0$ and the $f'(0)<0$ to show that $f(b)$ is decreasing for all $b>0$? $\endgroup$ – Boby Jun 15 '16 at 16:39
  • $\begingroup$ For $0<b<\delta$, where $\delta$ is small, $f(b)<0$ $\endgroup$ – HK Lee Jun 15 '16 at 16:42
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    $\begingroup$ Ok. Yes, by definition of the derivative. But how do you extend this to the statement \emph{for all} $b>0$? $\endgroup$ – Boby Jun 15 '16 at 16:49
  • $\begingroup$ If $b>0$, we do not know signs of $f(b)$ and $f'(b)$ Only for $b=0$, as in the above we can calculate easily. But for $b>0$, I have no idea for calculation $\endgroup$ – HK Lee Jun 15 '16 at 16:53
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    $\begingroup$ Ok. So, the result then holds only in vicinity of $b$ close to zero, right? $\endgroup$ – Boby Jun 15 '16 at 17:13

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