Determinant in inner product space 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.
 A: 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.
A: Here is more of a comment than a proof, actually a partial proof limited to 3d, where $\vec v_1, \vec v_2,$ and $\vec v_3\in \Bbb R^3$. The determinant of $A$ is the triple scalar product
$$\det A=\vec v_1\cdot(\vec v_2\times \vec v_3)$$
Using the Cauchy-Schwartz inequality
$$
|\det A|\leq|\vec v_1||\vec v_2\times \vec v_3|
$$
Using the Levi-Civita symbol and Einstein notation for the vector and scalar products
\begin{align}
|\vec v_2\times \vec v_3|^2&=\vec v_2\times \vec v_3\cdot\vec v_2\times \vec v_3=\epsilon_{ijk}v_{2j}v_{3k}\epsilon_{ilm}v_{2l}v_{3m}=
\epsilon_{ijk}\epsilon_{ilm}v_{2j}v_{3k}v_{2l}v_{3m}\\
&=v_{2j}v_{3k}v_{2j}v_{3k}-v_{2j}v_{3k}v_{2k}v_{3j}=v_2^2v_3^2-(\vec v_2\cdot\vec v_3)^2\leq v_2^2v_3^3,
\end{align}
where I used the identity $\epsilon_{ijk}\epsilon_{ilm}=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}$, in which (the Kronecker delta) $\delta_{jl}$ is 1 if $j=l$ and 0 otherwise.
Consequently, we obtain
$$|\det A|\leq v_1v_2v_3,$$
where $v_i=|\vec v_i|$.
