I am trying to tackle the following problem:
Given matrix $\alpha=(\alpha_{ij})$ such that all the entries are non-negative and $\alpha_{ij}\alpha_{ji}\leq 1$ for all $i,j$ and $\alpha_{ii}=1$ for all $i$. Find a vector $c=(c_1,c_2, \cdots , c_n)\geq 0$ such that $\alpha_{ji}\leq \frac{c_i}{c_j} \leq \frac{1}{\alpha_{ij}}$. (Clearly, this can have many solutions and we are just interested in one possible solution.)
I tried by assuming $c_1=1$ and use terms like $\max_{i}\alpha_{ij}$ but none of the attempts are successful. I would be thankful for any help!
Unfortunately, there is no such vector $c$ in general. A counterexample is the following. Let $\alpha$ be $3\times 3$ matrix such that $\alpha_{13}=\alpha_{32}=2$, $\alpha_{23}=\alpha_{31}=\frac 12$, and all remaining $\alpha_{ij}$ equal to $1$. Assume that $c$ is the required vector. Remark that since each $c_i$ can be placed in a denominator, all $c_i$ are non-negative. We have $1=\alpha_{21}\le\frac{c_1}{c_2}\le\frac {1}{\alpha_{12}}=1$, so $c_1=c_2=c’$. Also we have $\alpha_{32}c_3\le c’$ and $\alpha_{13}c’\le c_3$. That is $2c_3\le c’$ and $2c’\le c_3$. Hence $2c_3\le c’\le \frac 12 c_3$, a contradiction.