Let $M \in \mathbb{R}^{n \times n}$ be a symmetric matrix with positive entries and with dominant diagonal. This means that, for any $i$, $$M_{i,i} > \sum_{j\neq i} M_{i,j}.$$ Thanks to Hadamard's lemma, we know that $M$ is invertible.
Is it true that we always have that $$(M^{-1})_{i,i} > (M^{-1})_{i,j}$$ for any $i \neq j$?
Any help would be appreciated.
The is true. Partition $M$ as $$ \pmatrix{m_{11}&v^T\\ v&S}. $$ Using Schur complement, we obtain $$ N:=M^{-1}=\pmatrix{(m_{11}-v^TS^{-1}v)^{-1}&\ast\\ -(m_{11}-v^TS^{-1}v)^{-1}S^{-1}v&\ast}. $$ As $M$ is a strictly diagonally dominant real symmetric matrix with a positive diagonal, it is positive definite. Therefore $n_{11}=(m_{11}-v^TS^{-1}v)^{-1}>0$. It follows that $n_{11}>n_{i1}$ for all $i\ne 1$ if and only if $-S^{-1}v<e$, i.e., if and only if $S^{-1}v+e>0$.
Denote by $D$ and $F$ the diagonal and off-diagonal parts of $S$ respectively, so that $S=D+F$. Since $M$ is entrywise positive and strictly diagonally dominant, $(I\pm D^{-1}F)^{-1}$ can be expanded as a Neumann series, $(D-F)^{-1}$ is entrywise positive, and $(D-F)e-v>0$. It follows that $e>(D-F)^{-1}v$ and $$ \begin{aligned} S^{-1}v+e &=(D+F)^{-1}v+e\\ &>(D+F)^{-1}v+(D-F)^{-1}v\\ &=\left[(I+D^{-1}F)^{-1}+ (I-D^{-1}F)^{-1}\right]D^{-1}v\\ &=2\left[I+(D^{-1}F)^2+(D^{-1}F)^4+\ldots\right]D^{-1}v\\ &>0. \end{aligned} $$