I have recently made the following observation:


Let $v\_i := (v\_{i1}, v\_{i2})$, $1 \leq i \leq k$, be  non-zero   *positive*  elements of $\mathbb{Q}^2$ such that no two of them are proportional. Let $M$ be the $k \times k$ matrix whose entries are $m\_{ij} := \max${$v\_{ik}/v\_{jk}: 1 \leq k \leq 2$}. Then $\det M \neq 0$. 


The above statement is equivalent to the basic case of a result I recently discovered about pull back of divisors under a birational mapping of algebraic surfaces. I was going to include it as a part of another paper, then noticed the equivalent statement stated above and found it a bit amusing. My question is: is it worthwhile to try to publish it in a journal (as an example of an application of algebraic geometry to derive an arithmetic inequality), and if it is, then which journal(s)? 


It is of course also very much possible that it is already known, or has a trivial proof (or counterexample!) - anything along those directions would also be appreciated. 


 **Edit:**  Let me elaborate a bit about the geometric statement. In the 'other' paper, I define, for two algebraic varieties $X \subseteq Y$, something called "linking number at infinity" (with respect to $X$) of two divisors with support in $Y \setminus X$. I can show that when $Y$ is a surface, (under some additional conditions) the matrix of linking numbers at infinity of the divisors with support in $Y \subseteq X$ is non-singular. In a special (toric) case, the matrix of linking numbers takes the form of $M$ defined above. So the question is if the result about non-singularity of the matrix and its corresponding implication(s) are publishable anywhere.



Appropriate journal to publish a determinantal inequality
It seems to be true, but relatively simple (unless I made a mistake). Let's see:


First of all, scaling any pair $(v\_{i,1},v\_{i,2})$ by a constant $c$ does not change the determinant (one row of the matrix is multiplied by $c$, and one column is divided by $c$). We can therefore assume without losing generality that $(v\_{i,1},v\_{i,2})=(v\_i,1)$.


Also, permutation of $v\_i$'s does not change the determinant; therefore, we may assume that
$v\_i$'s are strictly increasing. In this case, the matrix has the form
$$m\_{ij}=\begin{cases} 1,& i\le j\cr v\_i/v\_j,& i>j\end{cases}.$$
Now subtract the top row from all others, and shift it to the bottom. You end up with a triangular matrix whose determinant is non-zero.


