If $E$ is an $n$-dimensional euclidean space then for all $u\in\Hom(E)$ and any orthonormal basis $e_1,\ldots,e_n$:
$$
|\det u|\leq\prod_j\norm{u(e_j)}~.
$$
$\proof$
1. For any $w\in\Hom(E)$ we have by the geometric-arithmetic mean inequality:
\begin{eqnarray*}
\det(w^2)^{1/n}
&=&\det(w^*w)^{1/n}
=(\l_1\cdots\l_n)^{1/n}
\leq\tfrac1n(\l_1+\cdots+\l_n)\\
&=&\tfrac1n\sum\la w^*w(x_j),x_j\ra
=\tfrac1n\sum\Vert w(x_j)\Vert^2
=\tfrac1n\Vert w\Vert_{HS}^2
\end{eqnarray*}
where $x_1,\ldots,x_n$ is an orthonormal basis of eigen vectors of $w^*w$ with eigen values $\l_j\geq0$.
2. Define $v\in\Hom(E)$ by $v:e_j\mapsto d_je_j$, then for any $u\in\Hom(E)$:
$$
(\det v^2\det(uu^*))^{1/n}
=(\det(uv)^2)^{1/n}
\leq\tfrac1n\norm{(uv)^2}_{HS}
=\tfrac1n\sum_j\norm{uv(e_j)}^2~.
$$
Now choose $d_j > 0$ such that $\prod d_j=1$ and $d_j\norm{u(e_j)}=d_k\norm{u(e_k)}$; by the equality case in the geometric-arithmetic mean inequality it follows that:
$$
\tfrac1n\sum_j\norm{uv(e_j)}^2
=\tfrac1n\sum_jd_j^2\norm{u(e_j)}^2
=\Big(\prod_j d_j\norm{u(e_j)}^2\Big)^{1/n}
=\Big(\prod_j \norm{u(e_j)}^2\Big)^{1/n}~.
$$
$\eofproof$