A rank-two counterexample to the subset inequality for the q-permanent

For $A=(a_{ij})\in\mathbb M_n$ and real $q$, let
$\operatorname{per}_q(A)=\sum_{\sigma}q^{\ell(\sigma)}\prod_i a_{i\sigma(i)}$,
where $\ell(\sigma)$ is the number of inversions, and let $R_{q,S}(A)$ be the
same sum restricted to permutations with $\sigma(S)=S$. The note shows that
the subset inequality $\operatorname{per}_q(A)\ge R_{q,S}(A)$, asked for
Hermitian positive semidefinite $A$ and $0\le q\le1$, is false. The
counterexample is a real positive semidefinite matrix of order four and rank
two, the Gram matrix $A=X^TX$ of an explicit integer $2\times4$ matrix $X$,
with $q=7/8$ and $S=\{2\}$. Expanding over the $24$ permutations gives
$\operatorname{per}_q(A)-R_{q,\{2\}}(A)=q(q+1)(115600q^4-229160q^3+315869q^2-344241q+145010)$,
which equals $-3235575/16384$ at $q=7/8$; all calculations are integer
polynomial arithmetic. By continuity the strict gap persists for
$A+\varepsilon I_4$ with small $\varepsilon>0$, so positive definite
counterexamples also exist.