This is an extension of Inequality involving matrix inverse elements, which has been already solved.
Let $A$ be an $N \times N$ matrix with all nonnegative entries and row sums strictly less than one, let $v$ be an $N$-dimensional vector with all nonnegative entries and weakly lower than one, let $D_0$ be an $n \times n$ nonnegative diagonal matrix such that $I \leq D_0 < I + \mathrm{diag} \left\{ \iota - A \iota \right\} \left( \mathrm{diag} \left\{ A \iota \right\} \right)^{-1}$ and where $I$ is an identity matrix and $\iota$ is a column-vector of all ones. Define $B \equiv \left(I - A \cdot \mathrm{diag}\{v \}\right)^{-1}$, $B_{1} \equiv \left(I - D_0 A \cdot \mathrm{diag}\{v \}\right)^{-1}$ and $B^{*}_{1} \equiv\left(I - D_0 A\right)^{-1}$. Here $\mathrm{diag}\left\{v\right\}$ is the diagonal matrix formed from vector $v$. I want to show that for any $i,j=1,...,N$ the following inequality holds: \begin{align*} v_{j} b_{ji} \dfrac{ b_{1,ji} }{ b^{*}_{1,ji} } + \sum_{k} \left(1-v_{k}\right) b_{ki} b_{jk} \dfrac{ b_{1,ki} }{ b^{*}_{1,ki} } \leq b_{ji}. \qquad (1) \end{align*}
If $D_{0} = I$ then we are in the case that was proved here.
We can prove (1) for $i = j$. For that, rewrite (1) for $i = j$ as \begin{align*} v_{i}b_{1,ii}+\sum_{k}\left(1-v_{k}\right)b_{ki}b_{ik}\dfrac{b_{1,ki}b_{1,ii}^{*}}{b_{ii}b_{1,ki}^{*}}\leq b_{1,ii}^{*}. \qquad (2) \end{align*} Using the same logic as here, we can show that $b_{1,ki}b_{ii}\geq b_{ki}b_{1,ii}$ for all $k$ and $i$. Indeed, we could show that $\sum_{k\neq i}d_{0,r}a_{rk}v_{k}g_{ki}\leq g_{ri}$, where $g_{ki}\equiv b_{1,ki}b_{ii}-b_{ki}b_{1,ii}$.
Also, since $D_{0}\geq I$ we have $I-D_{0}AV\leq I-AV$, which implies that $B_{1}\geq B$ or $b_{1,ik}\geq b_{ik}$ for all $i$ and $k$. Then we can write an upper bound for the left-hand side of (2): \begin{align*} v_{i}b_{1,ii}+\sum_{k}\left(1-v_{k}\right)\dfrac{b_{ki}b_{ik}}{b_{ii}}\cdot\dfrac{b_{1,ki}b_{1,ii}^{*}}{b_{1,ki}^{*}}\leq v_{i}b_{1,ii}+\sum_{k}\left(1-v_{k}\right)\dfrac{b_{1,ki}b_{1,ik}}{b_{1,ii}}\cdot\dfrac{b_{1,ki}b_{1,ii}^{*}}{b_{1,ki}^{*}}. \end{align*} Recylcing the proof from here, we have the inequality \begin{align*} v_{i}b_{1,ii}+\sum_{k}\left(1-v_{k}\right)\dfrac{b_{1,ki}b_{1,ik}}{b_{1,ii}}\cdot\dfrac{b_{1,ki}b_{1,ii}^{*}}{b_{1,ki}^{*}}\leq b_{1,ii}^{*}, \end{align*} which establishes inequality (2).
