Let $A$ be a matrix with all nonnegative entries and row sums strictly less than one, let $v$ be a vector with all entries between zero and one, and let $B\equiv\left(I-A\mathrm{diag}\left(v\right)\right)^{-1}$. Finally, consider function $f:\mathbb{R}^{N}\rightarrow\mathbb{R}$ given by $$ f(v)=v_{1}b_{11}+\sum_{k}\left(1-v_{k}\right)\frac{b_{1k}b_{k1}^{2}}{b_{11}}. $$ I want to show that $f(v)\leq f(\iota)$, where $\iota$ is the vector of all ones.
In the simple case in which $A$ is upper (lower) triangular then $B$ is also upper (lower) triangular, and hence $$ f(v)=\left[v_{1}+\left(1-v_{1}\right)b_{11}\right]b_{11}, $$ with $ b_{11}=\frac{1}{1-a_{11}v_{1}}. $ This implies that $f(v)\leq f(\iota)$ is equivalent to $$ \frac{1-a_{11}v_{1}^{2}}{\left(1-a_{11}v_{1}\right)^{2}}\leq\frac{1}{1-a_{11}}, $$ which is equivalent to $a_{11}\left(1-v_{1}\right)^{2}\geq0$, which is obviously true.
Extending this result beyond triangular matrices is difficult because now $\sum_{k}\left(1-v_{k}\right)\frac{b_{1k}b_{k1}^{2}}{b_{11}}\neq\left(1-v_{1}\right)b_{11}^{2}$ and also the expressions for $b_{ij}$ are complicated. Since numerical simulations show this result $f(v)\leq f(\iota)$ holds, I am hoping there is some trick that I can apply for this proof. Perhaps I would be using the set of equations in $B\left(I-A\mathrm{diag}\left(v\right)\right)=I$ and somehow use that to characterize $f(v)$.
Here is an answer for the special case in which $v_k = 1,\forall k\neq 1$. Let $B^{*}=\left(I-A\right)^{-1}$ and note that $f(\iota) = b_{11}^*$.The inequality can then be stated as $$v_{1}b_{11}+\left(1-v_{1}\right)b_{11}^{2}\leq b_{11}^{*}.$$
Let us use the fact that $b_{ij}=m_{ij}/\Delta$ and $b_{ij}^{*}=m_{ij}^{*}$/$\Delta^{*}$, where $M$ is the adjugate of $I-A\mathrm{diag}\left(v\right)$, $M^{*}$ is the adjugate of $I-A$ and $\Delta$ and $\Delta^{*}$ are the respective determinants. The inequality is then $$ v_{1}m_{11}/\Delta+\left(1-v_{1}\right)m_{11}^{2}/\Delta^{2}\leq m_{11}^{*}/{\Delta^{*}}. $$ Multiplying by $\Delta^{2}$ we get $$ v_{1}m_{11}\Delta+\left(1-v_{1}\right)m_{11}^{2}\leq\frac{m_{11}^{*}\Delta^{2}}{\Delta^{*}}. $$
Recall that $M$ is the adjugate of $I-A\mathrm{diag}\left(v\right)$, hence $$ m_{11} = \det \left[\begin{array}{ccccc} 1-a_{22}v_{2} & -a_{23}v_{3} & ... & ... & -a_{2K}v_{K}\\ -a_{32}v_{2} & 1-a_{33}v_{3} & ... & ... & -a_{3K}v_{K}\\ ... & ... & ... & ... & ...\\ ... & ... & ... & ... & ...\\ -a_{K2}v_{2} & -a_{K3}v_{3} & ... & ... & 1-a_{KK}v_{k} \end{array}\right] $$ and so $$m_{11} = \det \left[\begin{array}{ccccc} 1-a_{22} & -a_{23} & ... & ... & -a_{2K}\\ -a_{32} & 1-a_{33} & ... & ... & -a_{3K}\\ ... & ... & ... & ... & ...\\ ... & ... & ... & ... & ...\\ -a_{K2} & -a_{K3} & ... & ... & 1-a_{KK} \end{array}\right]=m_{11}^{*}.$$ Since $m_{11}^{*}>0$ then our inequality becomes $$ v_{1}\Delta+\left(1-v_{1}\right)m_{11}\leq\frac{\Delta^{2}}{\Delta^{*}} $$ Moreover, we know that \begin{align*} \Delta & =\left(1-a_{11}v_{1}\right)m_{11}-a_{12}v_{2}m_{21}...-a_{1K}v_{K}m_{K1}\\ & =\left(1-a_{11}v_{1}-1+a_{11}\right)m_{11}^{*}+\left(1-a_{11}\right)m_{11}^{*}-a_{12}m_{21}...-a_{1K}m_{K1}\\ & =\left(1-v_{1}\right)a_{11}m_{11}^{*}+\left(1-a_{11}\right)m_{11}^{*}-a_{12}m_{21}...-a_{1K}m_{K1} \end{align*} But $$ I-A\mathrm{diag}(v)=\left[\begin{array}{ccccc} 1-a_{11}v_{1} & -a_{12}v_{2} & -a_{13}v_{3} & ... & -a_{1K}v_{K}\\ -a_{21}v_{1} & 1-a_{22}v_{2} & -a_{23}v_{3} & ... & -a_{2K}v_{K}\\ -a_{31}v_{1} & -a_{32}v_{2} & 1-a_{33}v_{3} & ... & ...\\ ... & ... & ... & ... & ...\\ -a_{K1}v_{1} & -a_{K2}v_{2} & -a_{K3}v_{3} & ... & 1-a_{KK}v_{k} \end{array}\right] $$ so $$ m_{21}=-\det\left[\begin{array}{ccccc} -a_{21}v_{1} & -a_{23}v_{3} & ... & ... & -a_{2K}v_{K}\\ -a_{31}v_{1} & 1-a_{33}v_{3} & ... & ... & -a_{2K}v_{K}\\ ... & ... & ... & ... & ...\\ ... & ... & ... & ... & ...\\ -a_{K1}v_{1} & -a_{K3}v_{3} & ... & ... & 1-a_{KK}v_{k} \end{array}\right]. $$ In the case here with $v_{k}=1$ if $k\neq1$ we have \begin{align*} m_{21} & =-\det\left[\begin{array}{ccccc} -a_{21}v_{1} & -a_{23} & ... & ... & -a_{2K}\\ -a_{31}v_{1} & 1-a_{33} & ... & ... & -a_{2K}\\ ... & ... & ... & ... & ...\\ ... & ... & ... & ... & ...\\ -a_{K1}v_{1} & -a_{K3} & ... & ... & 1-a_{KK} \end{array}\right]\\ & =-v_{1}\det\left[\begin{array}{ccccc} -a_{21} & -a_{23} & ... & ... & -a_{2K}\\ -a_{31} & 1-a_{33} & ... & ... & -a_{2K}\\ ... & ... & ... & ... & ...\\ ... & ... & ... & ... & ...\\ -a_{K1} & -a_{K3} & ... & ... & 1-a_{KK} \end{array}\right]=v_{1}m_{21}^{*} \end{align*} and similarly $ m_{k1}=v_{1}m_{k1}^{*}, $ and so plugging above we get \begin{align*} \Delta & =\left(1-v_{1}\right)a_{11}m_{11}^{*}+\left(1-a_{11}\right)m_{11}^{*}-v_{1}\sum_{k\ge2}a_{1k}m_{k1}^{*}\\ & =\left(1-v_{1}\right)a_{11}m_{11}^{*}+\left(1-a_{11}\right)m_{11}^{*}-\sum_{k\ge2}a_{1k}m_{k1}^{*}+\left(1-v_{1}\right)\sum_{k\ge2}a_{1k}m_{k1}^{*}\\ & =\left(1-v_{1}\right)\sum_{k\ge1}a_{1k}m_{k1}^{*}+\Delta^{*}\\ & =\left(1-v_{1}\right)\Phi+\Delta^{*} \end{align*} where $$ \Phi\equiv\sum_{k\ge1}a_{1k}m_{k1}^{*}. $$ Plugging above then our inequality becomes $$ v_{1}\Delta+\left(1-v_{1}\right)m_{11}\leq\frac{\Delta^{2}}{\Delta^{*}} $$ $$ v_{1}\left(\left(1-v_{1}\right)\Phi+\Delta^{*}\right)+\left(1-v_{1}\right)m_{11}^{*}\leq\frac{\left(\left(1-v_{1}\right)\Phi+\Delta^{*}\right)^{2}}{\Delta^{*}} $$ $$ \left(1-v_{1}\right)v_{1}\Phi+\Delta^{*}v_{1}+\left(1-v_{1}\right)m_{11}^{*}\leq\frac{\left(1-v_{1}\right)^{2}\Phi^{2}+2\left(1-v_{1}\right)\Phi\Delta^{*}+\Delta^{*2}}{\Delta^{*}} $$ $$ \left(1-v_{1}\right)v_{1}\Phi+\left(1-v_{1}\right)m_{11}^{*}\leq\frac{\left(1-v_{1}\right)^{2}\Phi^{2}}{\Delta^{*}}+2\left(1-v_{1}\right)\Phi+\left(1-v_{1}\right)\Delta^{*} $$ For $0\leq v_{1}<1$ this is equivalent to $$ v_{1}\Phi+m_{11}^{*}\leq\frac{\left(1-v_{1}\right)\Phi^{2}}{\Delta^{*}}+2\Phi+\Delta^{*} $$ or $$ \left(\left(v_{1}-2\right)\Phi+m_{11}^{*}\right)\Delta^{*}\leq\left(1-v_{1}\right)\Phi^{2}+\left(\Delta^{*}\right)^{2} $$ But $$ \Delta^{*}=\left(1-a_{11}\right)m_{11}^{*}-\sum_{k\ge2}a_{1k}m_{k1}^{*} $$ and $$ \Phi\equiv\sum_{k\ge1}a_{1k}m_{k1}^{*} $$ so $$ \Delta^{*}=m_{11}^{*}-\Phi $$ and hence our inequality is $$ \left(\left(v_{1}-2\right)\Phi+m_{11}^{*}\right)\left(m_{11}^{*}-\Phi\right)\leq\left(1-v_{1}\right)\Phi^{2}+\left(m_{11}^{*}-\Phi\right)^{2} $$ or $$ \left(v_{1}-2\right)m_{11}^{*}\Phi+\left(m_{11}^{*}\right)^{2}-\left(v_{1}-2\right)\Phi^{2}-m_{11}^{*}\Phi\leq\left(1-v_{1}\right)\Phi^{2}+\left(m_{11}^{*}\right)^{2}-2m_{11}^{*}\Phi+\Phi^{2} $$ or $$ \left(v_{1}-1\right)m_{11}^{*}\Phi-\left(v_{1}-2\right)\Phi^{2}\leq\left(1-v_{1}\right)\Phi^{2}+\Phi^{2} $$ or $$ \left(v_{1}-1\right)m_{11}^{*}\Phi-v_{1}\Phi^{2}+2\Phi^{2}\leq2\Phi^{2}-v_{1}\Phi^{2} $$ or $$ -\left(1-v_{1}\right)m_{11}^{*}\Phi\leq0, $$ which is true.
I imagine one can work this out in the case which $v_2$ can also be lower than one but $v_k = 1,\forall k \neq 1,2$, and from there try to come up with an induction argument to prove the general case.
