Suppose $A=(a_{ij})$ is a symmetric (0,1)-matrix with 1's along the diagonal, and let $A_{ij}$ be the matrix obtained by removing the $i$-th row and $j$-th column.

Based on substantial numerical experimentation it seems that the bound $$\operatorname{Perm}(A) \geq 2 \operatorname{Perm}(A_{ij})$$ holds for any $(i,j)$, $i \neq j$. I would be interested in any bound of the form $\operatorname{Perm}(A) \geq c \operatorname{Perm}(A_{ij})$ for a constant $c>0$. (The bound $\operatorname{Perm}(A) \geq \operatorname{Perm}(A_{ij})$ is trivial if $a_{ij}=1$, but not so clear if $a_{ij}=0$.)

My numerical experimentation suggests that this is a specific case of a far more general fact:

**Conjecture:** If $A$ is any matrix with nonnegative entries, $i \neq j$ and $\operatorname{Perm}(A)>0$, then $$\operatorname{Perm}(A)^2 \geq 4 a_{ii}a_{jj} \operatorname{Perm}(A_{ij})\operatorname{Perm}(A_{ji}). $$

It seems that there are various families of matrices $A$ that achieve equality.