I have some non-negative real variables $x_1, \ldots ,x_n$ and non-negative real constants $a_{i,j}, 1 \le i \le n, 1 \le j \le n$, such that $a_{i,j} = 0 \iff i > j+1$, and other positive real constants $b_k, 1 \le k \le n$ such that:
$$\begin{align} a_{1,1}x_1-a_{1,2}x_2+\cdots +(-1)^{1+n}a_{1,n}x_n & \ge b_1 \\ -a_{2,1}x_1+a_{2,2}x_2+\cdots +(-1)^{2+n}a_{2,n}x_n & \ge b_2 \\ & \cdots \\ -a_{n-1,n-2}x_{n-2}+a_{n-1,n-1}x_{n-1}-a_{n-1,n}x_n & \ge b_{n-1} \\ -a_{n,n-1}x_{n-1}+a_{n,n}x_n & \ge b_n \end{align} $$
Each $a_{i,j} \not= 0$ is multiplied by $(-1)^{i+j}$.
I believe that it is always possible to combine the inequalities to get:
$$x_1 \ge A(a_{1,1}, \ldots ,a_{n,n}, b_1, \ldots , b_n)$$
where $A$ is a constant, function of all the $a_{i,j}$ and $b_k$ only. My goal is getting an analytical formula for $A$, if actually possible.
I am able to do it for some specific small cases, but I suspect that there might be some well known systematic way of doing it, maybe involving determinants, in a way perhaps similar to solving linear systems of equations.
I read about Fourier-Motzkin elimination, however I would like a closed form expression, and I think my case is simpler than the general setup because I have a square $n \times n$ "nearly" triangular matrix made by the $a_{i,j}$ coefficients.
Any hint?
EDIT 2023-01-15: I have now asked a more precise version of this question at mathoverflow.
If you solve a linear programming problem that minimizes $x_1$, yielding an optimal solution with $x_1=A$, the resulting dual variables provide a certificate of optimality. Explicitly, multiply constraint $i$ by the dual variable $y_i$ and then add everything up to derive valid inequality $x_1 \ge A$.