Showing $ u^T (M+I_n)^{-1} u \geq v^T(M+I_n)^{-1}v $ when $M$ is symmetric PD with positive entries and $u,v$ are $0-1$ vectors

Question

Let $n$ be a positive integer. Let $m_u,m_v \in \{1,...,n-1 \}$.

Let $M$ be a $n \times n$ symmetric positive definite matrix with positive entries.

Let $u$ and $v$ be vectors of length $n$ with entries consisting $n-m_u$ (or $n-m_v$) $0$'s and $m_u $ (or $m_v$) $1$'s. Sort $u$ so that the first $n-m_u$ entires of $u$ are $0$'s and the last $m_u$ entries are $1$'s. Sort $v$ in the same way.

Suppose $m_u>m_v$. Is the following weak inequality true?

(As shown below by @Niki Di Giano, this is not true)

$$ u^T (M+I_n)^{-1} u \geq v^T(M+I_n)^{-1}v $$

This is related to the post but now $M$ has positive entries only and we replace $M$ with $(M+I_n)^{-1}$.

Answer 1

Using almost the same matrix as the other post (if the matrix admits zero entries): $$M = \begin{bmatrix} 1 & 0 & 0\\ 0 & 10 & 9\\ 0 & 9 & 10\\ \end{bmatrix} $$ Then its inverse with the added identity is: $$(M + I)^{-1}= \begin{bmatrix} 1/2 & 0 & 0\\ 0 & 11/40 & -9/40\\ 0 & -9/40 & 11/40\\ \end{bmatrix} $$ This gives $u^T M u = 4/40$ and $v^T M v = 11/40$ for the vectors $u = (0, 1, 1)$ and $v =(0, 0, 1)$ respectively.

Alternatively, the matrix:

$$M = \begin{bmatrix} 1 & 1 & 1\\ 1 & 100 & 99\\ 1 & 99 & 100\\ \end{bmatrix} \\ \implies (M + I)^{-1} = \begin{bmatrix} 100/199 & -1/398 & -1/398\\ -1/398 & 201/796 & - 197/796\\ -1/398 & - 197/796 & 201/796\\ \end{bmatrix} $$ Which gives $8/796$ and $201/796$ for the same two vectors.