Let $A$ be a $n\times n$ matrix, then the permanent of $A$ is defined as $$per A=\sum_{\sigma} a_{1\sigma(1)}\cdots a_{n\sigma(n)}.$$ Now, it is to show that $$|per(AB)|^2\leq per (AA^*)per(B^*B).$$
I find it is difficult to obtain by using the Cauchy-Schwarz inequality.