Hadamard's inequality

Geometric interpretation. $|\det A|$ is the volume of the parallelotope spanned by $C_1,\ldots,C_n$. This volume equals the product of the lengths of the columns *only* when the columns are orthogonal. In general, the volume is at most the product of lengths (the "base times height" formula, with the height bounded by the corresponding side).

Formal proof by Gram-Schmidt. Apply Gram-Schmidt to $C_1,\ldots,C_n$: let $q_1,\ldots,q_n$ be the orthonormal basis, and $r_{ij}$ the coefficients. Then $A=QR$ with $R$ upper triangular, and $|\det A|=|\det Q||\det R|=\prod_i|r_{ii}|$. But $|r_{ii}|\le\|C_i\|$ (each $r_{ii}$ is the length of the component of $C_i$ orthogonal to the previous columns, hence bounded by $\|C_i\|$).

Conclusion. $|\det A|\le\prod\|C_i\|$, with equality iff the columns are orthogonal.