Let $A\in\mathbb{R}^{n\times n}$ be a symmetric matrix with $A_{i,i} = 1$ and $A_{i,j}\in[1/2,1]$ for every $i\neq j$. Let $v = (1,\dots,1)\in\mathbb{R}^n$ be the all ones vectors. I need to show that:
$v^\top A^{-1}v \leq 2$.
For the case where $A_{i,j}=\frac{1}{2} + c$ for some $c\in[0,1/2]$, $v$ is an eigevector with an eigenvalue of $\alpha:= 1 + \frac{n-1}{2} + (n-1)c$. In this case, since $A$ is symmetric, $v$ is also an eigenvector of $A^{-1}$ with an eigenvalue of $\alpha^{-1}$, hence $v^\top A^{-1}v=\alpha^{-1}v^\top v = n\alpha^{-1} \leq 2$ for any value of $c$. My problem is when the off-diagonal entries are not all equal hence $v$ is not necessarily an eigenvector of $A$.