Suppose $M$ is a real positive definite matrix, and $\Lambda$ is a nonnegative diagonal matrix, then does the following inequality always hold? $$(M^{-2})_{ii} - ((M + \Lambda)^{-2})_{ii} \ge 0$$ where $(\cdot)_{ii}$ denotes the $(i,i)$-element, and $M$ satisfies that:
1). All the off-diagonal elements of $M$ are $\le 0$;
2). $M$ has its row sums all $ > 1$, or in notations: $$(M)_{ii} +\sum_{j\ne i} (M)_{ij}> 1$$ (If you want more context, $M$ is a laplacian matrix plus a positive diagonal matrix.)
Intuitively $(M + \Lambda)$ should be "more positive definite" than $M$, and $(M+\Lambda)^{-2}$ "less positive definite" than $M^{-2}$... but I don't know how to start at all.
The assumptions on $M$ can be relaxed. Forget the row sums of $M$. Suppose $M$ is symmetric positive definite and it has non-positive off-diagonal entries. Then $M+D$ is an M-matrix for any nonnegative diagonal matrix $D$. Hence $(M+D)^{-1}>0$ entrywise.
Now, if $M$ is an M-matrix and $D$ is a nonnegative diagonal matrix with exactly one nonzero (and positive) diagonal entry, then $M^{-1}>(M+D)^{-1}$ entrywise. This is a direct consequence of Sherman-Morrison formula (and the fact that $M^{-1}$ is both entrywise positive and positive definite).
So, if $\Lambda$ is a nonnegative diagonal matrix, by applying Sherman-Morrison formula repeatedly, we get $M^{-1}\ge(M+\Lambda)^{-1}$ entrywise (the inequality is actually strict if $\Lambda$ is nonzero). In turn, $M^{-2}=(M^{-1})^2\ge\left(M+\Lambda)^{-1}\right)^2=(M+\Lambda)^{-2}$.