Why does $\left| {\det A} \right| \le \prod\limits_{j = 1}^n {(\sum\limits_{i = 1}^n {\left| {{a_{ij}}} \right|} )} $

Question

Let $A \in {M_n}(\mathbb{C})$ and $A = \left[ {{a_{ij}}} \right]$.

Why does $\left| {\det A} \right| \le \prod\limits_{j = 1}^n {(\sum\limits_{i = 1}^n {\left| {{a_{ij}}} \right|} )} $ is true?

Answer 1

By definition, $$\det A=\sum_{(i_1\cdots i_n)}(-1)^{\text{sgn}(i_1\cdots i_n)}a_{1i_1}\cdots a_{ni_n}.$$ Hence, $$|\det A|\le\sum_{(i_1\cdots i_n)}|a_{1i_1}|\cdots |a_{ni_n}|.$$ Verify that $$\prod\limits_{j = 1}^n {(\sum\limits_{i = 1}^n {\left| {{a_{ij}}} \right|} )}\ge\sum_{(i_1\cdots i_n)}|a_{1i_1}|\cdots |a_{ni_n}|$$ and complete the proof. (Note that there are $n^n$ terms on the left while $n!$ terms on the right.)