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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 form $L_i(\underline X)=\sum_{i=1}^ma_ix_i$ 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

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

In the book Number Theory IV from Parshin ( https://www.springer.com/gp/book/9783540614678 ), one can find this (page 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 form $L_i(\underline X)=\sum_{i=1}^ma_ix_i$ 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

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 form $L_i(\underline X)=\sum_{i=1}^ma_ix_i$ 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

In the book Number Theory IV from Parshin ( https://www.springer.com/gp/book/9783540614678 ), one can find this (page 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 form $L_i(\underline X)=\sum_{i=1}^ma_ix_i$ are linealy 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 proves that? Thanks in adavce

In the book Number Theory IV from Parshin ( https://www.springer.com/gp/book/9783540614678 ), one can find this (page 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 form $L_i(\underline X)=\sum_{i=1}^ma_ix_i$ 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