Let $U = [u_1 | \ldots | u_n]$ be a special unitary matrix (meaning that $u_i$ is its $i$-th column) and let $A$ be an $n$-by-$n$ hermitian matrix, where $n \geq 3$. I am interested in the following "Cauchy-Schwarz-like" inequality involving the determinants of $U$ whose first three columns have been acted upon by $A$ in a specific way: $$|\det [Au_1 | Au_2 | A^2u_3 | u_4 | \ldots | u_n]|^2 \\ \leq \det [A^2u_1 | u_2 | A^2u_3 | u_4 | \ldots | u_n] \cdot \det [u_1 | A^2u_2 | A^2u_3 | u_4 | \ldots | u_n]$$ or, if you prefer, $$\det \left[\begin{array}{cc} \det [A^2u_1 | u_2 | A^2u_3 | u_4 | \ldots | u_n] & \det [Au_1 | Au_2 | A^2u_3 | u_4 | \ldots | u_n] \\ \overline{\det [Au_1 | Au_2 | A^2u_3 | u_4 | \ldots | u_n]} & \det [u_1 | A^2u_2 | A^2u_3 | u_4 | \ldots | u_n] \end{array} \right] \geq 0.$$ It holds trivially (and saturates) if $u_1, u_2, u_3$ are eigenvectors of $A$, but how to prove it in the general case?
EDIT: Actually, without loss of generality one can assume that $U = I_n$.