Looking at the vector space $\Bbb C^n$ with the standart inner product, Let $v_1,...v_n \in \Bbb C^n$ and $A \in M_n(\Bbb C)$ a matrix with columns $v_1,...,v_n$. Prove that:
$$ \lvert detA\rvert \leq \prod_{j=1}^n ||v_j||$$
Furthermore, prove that equality holds if and only if $v_1,...,v_n$ is an orthogonal sequence.
If $\operatorname{rank} A < n$ then $\det(A) = 0$ and the inequality is clear. In this case, the equality holds iff $v_i = 0$ for some $1 \leq i \leq n$ (note that this doesn't mean that $v_1,\dots,v_n$ is an orthogonal sequence).
On the other hand, if $A$ has full rank, perform a QR decomposition and write $A = QR$ when $Q$ is unitary and $R$ is upper triangular. The columns $e_1,\dots,e_n$ of $Q$ are the result of performing the Gram-Schmidt procedure on $v_1,\dots,v_n$ while the diagonal entries of $R$ are $\left< v_i, e_i \right>$. Hence,
$$ |\det(A)| = |\det(QR)| = |\det(Q)||\det(R)| = |\det(R)| = \prod_{i=1}^n |\left< v_i, e_i \right>| \leq \prod_{i=1}^n \| v_i \| \| e_i \| = \prod_{i=1}^n \| v_i \|. $$
Equality holds if and only if $| \left< v_i, e_i \right> | = \| v_i \|$ for all $1 \leq i \leq n$. Since equality in the Cauchy-Schwartz inequality holds if and only if the vectors are linearly dependent, we can write $v_i = c_i e_i$ for some $c_i \in \mathbb{C}$ with $\| v_i \| = c_i$ which implies that $(v_i)_{i=1}^n$ are orthogonal because $(e_i)_{i=1}^n$ are.