Inequality involving traces and Toeplitz matrices Suppose that I have $n\times n$ matrices $\mathbf{A}$, $\mathbf{B}$ and $\mathbf{B}_{\epsilon}$ that are real with non-negative entries. The following properties hold for these matrices:


*

*$\mathbf{A}$ is a symmetric Toeplitz matrix.
Denote by $a_d$ the entries in the diagonal $d$ away from the main (central) diagonal. 
For some constant $T<n$, the entries $a_t=0$ for all $t\geq T$ and
$a_t>0$ for all $t<T$.  Thus, $\mathbf{A}$ has the following form:
$$\mathbf{A}=\overbrace{\left.\left[\begin{array}{11c}
   a_1&a_2&a_3&\cdots&a_t&0&0&\cdots&0&0&0\\
   a_2&a_1&a_3&\cdots&a_{t-1}&a_t&0&\cdots&0&0&0\\
   a_3&a_2&a_1&\cdots&a_{t-2}&a_{t-1}&a_t&\cdots&0&0&0\\
   &\vdots&&&&&&&&\vdots&\\
   0&0&0&\cdots&a_{t}&a_{t-1}&a_{t-2}&\cdots&a_1&a_2&a_3\\
   0&0&0&\cdots&0&a_t&a_{t-1}&\cdots&a_2&a_1&a_2\\
   0&0&0&\cdots&0&0&a_t&\cdots&a_3&a_2&a_1\\
   \end{array}\right]\right\}}^nn$$
(in my problem, $\mathbf{A}$ is a covariance matrix.)

*$\mathbf{B}$ is a diagonal matrix with entries equal to $b>0$:  $\mathbf{B}=b\mathbf{I}$, where $\mathbf{I}$ is the identity matrix.

*$\mathbf{B}_{\epsilon}$ is a diagonal matrix defined as follows: given constants $\epsilon_0$ and $\epsilon_1$ such that $0<\epsilon_0<b$ and $\epsilon_0<\epsilon_1$, set two arbitrary entries $B_\epsilon[i,i]$ and $B_\epsilon[j,j]$ ($i\neq j$) in the diagonal of $\mathbf{B}_{\epsilon}$ to $b-\epsilon_0$ and $b+\epsilon_1$, and set the rest of the diagonal entries to $b$.

*$\mathbf{A}+\mathbf{B}$ and $\mathbf{A}+\mathbf{B}_\epsilon$ are invertible.


I am interested in finding out whether asymptotically with $n\rightarrow\infty$ for constant $T$, $\epsilon_0$ and $\epsilon_1$ and the listed conditions:
$$\operatorname{Tr}[\mathbf{A}(\mathbf{A}+\mathbf{B})^{-1}\mathbf{A}]\geq\operatorname{Tr}[\mathbf{A}(\mathbf{A}+\mathbf{B}_{\epsilon})^{-1}\mathbf{A}]$$
where $\operatorname{Tr}[\mathbf{M}]$ denotes the trace of the matrix $\mathbf{M}$.
I have a hunch that it's true and that there exists $n_0$ such that for all $n>n_0$ the statement above is true.  This hunch is based on the fact that $\operatorname{Tr}[\mathbf{A}+\mathbf{B}]\leq\operatorname{Tr}[\mathbf{A}+\mathbf{B}_\epsilon]$ and that we are "dividing" by that (I know that the trace of the product is not the product of the trace, hence "dividing" in quotations).  However, my knowledge of linear algebra is not sufficient to prove this and my hunch may be wrong.  Any help would be appreciated.

In the simplest example, your condition fails: let $A=I$, $b=1$, $\epsilon_1\in[0,1]$, $\epsilon_0=\epsilon_1/2$. Then $A+B=2I$, so
$$
\operatorname{Tr}(A(A+B)^{-1}A)=\frac{n}2.
$$
Also, 
$$
\operatorname{Tr}(A(A+B_\epsilon)^{-1}A)=\frac{n-2}2+\frac1{2-\epsilon_0}+\frac1{2+\epsilon_1}.
$$
So
\begin{eqnarray}
\operatorname{Tr}(A(A+B)^{-1}A)-\operatorname{Tr}(A(A+B_\epsilon)^{-1}A)&=&-1+\frac1{2-\epsilon_0}+\frac1{2+\epsilon_1}=-1+\frac{4+\epsilon_1/2}{4+\epsilon_1-\epsilon_1^2/2}\\
&=&-\frac{\epsilon_1/2-\epsilon^2/2}{4+\epsilon_1-\epsilon_1^2/2}<0
\end{eqnarray}
