Minoration of linear forms In the book Number Theory IV from Parshin, one can find this statement (precisely at p. 215) with the comment "it is easy to see that...":

Let $a_{ij}$ ($1\le i,j\le m$) be complex numbers with $|a_{i,j}|\le H_i$ ($1\le j\le m$).
One assumes that the linear forms $L_i(\underline X)=\sum_{i=1}^ma_ix_i$ are linearly independent. Then, for any $\overline w=(w_1,\cdots,w_m)\in\mathbb C^m$, one has
$$\sum_{i=1}^m\frac{|L_i(\overline w)|}{H_i}\ge c_1\frac{|\Delta|}{H_1\cdots H_m}$$
where $\Delta=\det(a_{i,j})$ and $c_1=\frac1{m!}\max_{1\le j\le m}|w_j|$.

Unfortunately, for me it is not easy. Can one have hints to prove that? Thanks in advance
 A: We claim that
$$
\sum_{i=1}^m \frac{|L_i(w)|}{H_i} \geqslant \frac{|w|_\infty}{(m-1)!}\frac{|\Delta|}{H_1 \cdots H_m},
$$
where $|w|_\infty = \max_{1 \leqslant j \leqslant m} |w_j|$.
Indeed, first assume that $|A|_\infty = \max_{1\leqslant i,j \leqslant m} |a_{ij}|\leqslant 1$, and $H_i = 1$ for all $i$. Because $|Bw|_\infty \leqslant |B|_\infty |w|_1$ for any matrix $B$ and $w \in \mathbb{C}^m$ (where $|w| = \sum_{j=1}^m |w_j|$), we have
$$
|w|_\infty \leqslant |A^{-1}|_\infty|Aw|_1.
$$
Now $A^{-1} = \frac{1}{|\Delta|} \mathrm{Com}(A)^\top$ where $\mathrm{Com}(A)$ is the comatrix of $A$. Note that $|\mathrm{Com}(A)|_\infty \leqslant (m-1)!$ by the Leibniz formula for determinants, since $|A|_\infty \leqslant 1$. Thus
$$
|A^{-1}|_\infty \leqslant \frac{(m-1)!}{|\Delta|}.
$$
Therefore, by definition of $L_i$ $(i = 1, \dots, m)$ we get
$$
\sum_{i=1}^m |L_i(w)| = |Aw|_1 \geqslant \frac{|w|_\infty |\Delta|}{(m-1)!},
$$
which is the sought result in the case $|A|_\infty \leqslant 1$ and $H_1 = \cdots = H_m = 1$.
Finally take a general $A$ and assume that $\max_{1 \leqslant j \leqslant m}|a_{ij}| \leqslant H_i$ for $i = 1, \dots, m$. Writing $A = \begin{pmatrix} L_1 \\ \vdots \\ L_m \end{pmatrix}$ and applying the above estimate to the matrix
$
\tilde A = \begin{pmatrix} L_1 / H_1 \\ \vdots \\ L_m / H_m \end{pmatrix},
$
we get the claim.
