Minimal eigenvalue inequality My problem is to show that $$\lambda_{\min}(PA) \leq  \lambda_{\min}((D-M)A) ,$$ where $A$ is an arbitrary $n\times n$ symmetric positive definite matrix and $P$ is a diagonal matrix with $\frac{1}{A_{ii}}$ as the $i$-th diagonal element and $D$ is a diagonal matrix with the $i$-th diagonal element equal to $\sum_{j = 1}^n\frac{p_{ij}A_{jj}}{\det_{ij}}$ and $M$ is a symmetric matrix with $M_{ij} = \frac{p_{ij}A_{ij}}{\det_{ij}}$, where $p_{ij}$ are probabilities with $p_{ii} = 0,\, \forall i$,  $p_{ij} = \frac{1}{n-1}$ for $i\neq j$, and $\det_{ij} = A_{ii}A_{jj} - A_{ij}^2$ for $i \neq j$, and $\det_{ii} = 1$ for all $i$. $A_{ij}$ is the element on the $i$-th row and $j$-th column. I have tried many simulations and this statement always holds, even with some special cases $2$ can be substituted by arbitrary large constant, but I am not able to contstruct a proof except when $A$ is diagonal. 
 A: The following is an exploration. This gives a proof for $n=2$ and block diagonal $A$ with $2\times2$ blocks. Hopefully, someone will be inspired to find a proof for the general case.

$$A= \frac1{n-1}\sum_{i<j}A_{i,j},\quad D-M=\frac{1}{n-1}\sum_{i<j}A_{i,j}^{(-1)},\quad P=\frac1{n-1}\sum_{i<j} \big(\text{diag}(A_{i,j})\big)^{-1}$$
where $A_{i,j}$ is the $n\times n$ matrix with entry $(i,j)$ equal to entry $A_{ij}$, entry $(i,i)$ equal to $A_{ii}$ and entry $(j,j)$ equal to $A_{jj}$, and all other entries zero; where $A^{(-1)}_{i,j}$ is the $n\times n$ matrix the entries $(i,i),\,(i,j),\,(j,i),\,(j,j)$ of which equal to the corresponding entries of the inverse of the row $i$ column $j$ principal minor of $A$, and all other entries zero.
When $n=2$, $PA=\begin{bmatrix}1 & * \\
* & 1\end{bmatrix}$ which is positive definite. Its minimal eigvenvalue of is no larger than $1$ which is that minimal eigenvalue of its diagonal matrix $I$. $(D-M)A=I$. So the desired inequality holds. 
When $n>2$, 
$$(D-M)A=\frac1{n-1}\Big(I+\frac1{n-1}\sum_{i<j,\,k<l}A_{i,j}^{-1}A_{k,l}\Big),$$
where exactly two of $i,j,k,l$ are equal. 
