Is the conjecture $Per A $ is the largest eigenvalue of $\tilde{A}$ being solved?

Question

Let A be a positive semidefinite matrix of order $n$.

Is the conjecture $Per A $ is the largest eigenvalue of $\tilde{A}$ is being solved?

Where

1. $\tilde{A}$ is the matrix of order $n!\times n!$whose $(\sigma, \tau)^{th}$ entry is given by $\prod _{i=1}^{n} a_{{\sigma (i)}\tau(i)}$
2. $Per (A)=\sum _{\sigma\in S_n} \prod_{i=1}^{n} a_{i\sigma (i)}$

The conjecture has being proved for $n\leq 3$ in Bapat, R. B. ; Sunder, V. S. An extremal property of the permanent and the determinant. Linear Algebra Appl. 76 (1986), 153–163.

Is there any further development in this problem?

Answer 1

The permanent-on-top conjecture stating that per(A) is the largest eigenvalue of the Schur power matrix is false! See a counterexample for n=5 and rank = 2 here:

https://www.researchgate.net/publication/280495158_The_permanent_on_top_conjecture_is_false