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Suppose that $x$, $y$, $z$, $\nu$, $N$, $g$ are some postitive parameters and $A$, $B$ and $C$ are variables that belong to $(0,+\infty)$. I want to identify one or all of the feasible points taht solve the following system of inequalities with respect to $A$, $B$ and $C$

\begin{equation}xA-\nu gB-Ng C>0 \\ -\nu gA+yB-Ng C>0\\ -\nu gA-\nu gB +zC>0\end{equation}

Gary
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Nav89
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    By solve I suppose you mean to identify one or all of the feasible points. I'm not sure if you meant to make those inequalities strict. Perhaps if you provide some context for the Question (e.g. how did this problem arise and/or how you tried to solve it) then Readers can respond more easily with helpful information. – hardmath Feb 08 '21 at 18:26
  • Yes, I mean what you say! And let me change it. And yes these inequalities need to be strict. I do not know how to solve it exactly. Some idea could be to solve it as $3x3$ system with linear algebra...however i am not sure how to do so. Also, there is a chance to solve it with linear programming – Nav89 Feb 08 '21 at 18:36
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    It would be a standard linear program if these were strict inequalities. So I'd use a simplex algorithm for such a small system. The simplex method allows us to find the extremal points (vertices) of the feasible region (if there are any). By requiring strict inequalities all around, you are asking for the interior of the feasible region. – hardmath Feb 08 '21 at 18:44
  • I have never used simplex method. I need some help and this is why i asked for a help. Maybe the problem is simple for someone who knows the medthod, but for me it is not. Thank you in advance for the guidance. I am looking forward for any other help! – Nav89 Feb 08 '21 at 18:48
  • @hardmath what should i do to find the interior of the feasible region? I mean what method do i need to use now? – Nav89 Feb 08 '21 at 19:22
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    I'll write up an Answer for you. The special case you want to solve lends itself to a visual approach. It would still be helpful to future Readers to learn what motivates this problem, and I will upvote it if you add that. – hardmath Feb 08 '21 at 19:55
  • I will be glad if you do so! Of course I will add it! Thank you! – Nav89 Feb 08 '21 at 19:56
  • Is there a typo in the third inequality? The coefficient $\nu g$ appears there twice, in contrast to the first two inequalities. – hardmath Feb 09 '21 at 00:05

2 Answers2

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Too long for a comment.

Solving the system for $A,B,C$

$$ \left\{ \begin{array}{l} A x-B g \nu -C g N=e_1^2 \\ -A g \nu +B y-C g N=e_2^2 \\ -A g \nu -B g \nu +C z=e_3^2 \\ \end{array} \right. $$

we get

$$ \left\{ \begin{array}{c} A = \frac{e_1^2 \left(g^2 N \nu -y z\right)-g \left(e_3^2 N (g \nu +y)+e_2^2 \nu (g N+z)\right)}{g^2 N \nu (2 g \nu +x+y)+z \left(g^2 \nu ^2-x y\right)} \\ B = \frac{e_2^2 \left(g^2 N \nu -x z\right)-e_3^2 g N (g \nu +x)-e_1^2 g \nu (g N+z)}{g^2 N \nu (2 g \nu +x+y)+z \left(g^2 \nu ^2-x y\right)} \\ C= \frac{e_3^2 \left(g^2 \nu ^2-x y\right)-e_2^2 g \nu (g \nu +x)-e_1^2 g \nu (g \nu +y)}{g^2 N \nu (2 g \nu +x+y)+z \left(g^2 \nu ^2-x y\right)} \\ \end{array} \right. $$

or assuming $g^2 N \nu (2 g \nu +x+y)+z \left(g^2 \nu ^2-x y\right)>0$

$$ \left\{ \begin{array}{c} A' = e_1^2 \left(g^2 N \nu -y z\right)-g \left(e_3^2 N (g \nu +y)+e_2^2 \nu (g N+z)\right) \\ B' = e_2^2 \left(g^2 N \nu -x z\right)-e_3^2 g N (g \nu +x)-e_1^2 g \nu (g N+z) \\ C'= e_3^2 \left(g^2 \nu ^2-x y\right)-e_2^2 g \nu (g \nu +x)-e_1^2 g \nu (g \nu +y) \\ \end{array} \right. $$

NOTE

The matrix associated to the linear system is

$$ M = \left( \begin{array}{ccc} x & -g \nu & -g N \\ -g \nu & y & -g N \\ -g \nu & -g \nu & z \\ \end{array} \right) $$

and as long as $\det\left(M\right)=x y z-2 g^3 N \nu ^2-g^2 \nu (N (x+y)+\nu z)\ne 0$ we have $e_1, e_2, e_3$ as independent variations.

Cesareo
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  • what these $e_1$, $e_2$ and $e_3$ variables stand for...they transform the inequalites to equalities arent they? And why? – Nav89 Feb 08 '21 at 20:05
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    Note that not only the right hand side terms should be positive, but also the unknowns $A,B,C$ are required to be positive. – hardmath Feb 08 '21 at 20:15
  • Yes exactly! $A$ and $B$ and $C$ need also to be positive...their solutions – Nav89 Feb 08 '21 at 20:22
  • @Nav89 Here the intention was to solve for $A,B,C$ and also analyze under which circumstances $A,B,C$ are positive. $e_1,e_2,e_3$ were introduced to handle equations instead of inequalities. – Cesareo Feb 08 '21 at 20:46
  • @Cesareo you mean that this solution ensures that $A$, $B$ and $C$ are positive....? – Nav89 Feb 08 '21 at 20:56
  • Nope. In the expressions for $A',B',C'$ we can observe that the coefficients of $e_1^2,e_2^2,e_3^2$ have diverse signs, and if $\det(M)\ne 0$ those parameters are independent. Concluding, in the general case, with positive values for $x,y,\cdots,g$ scarcely the solution for $A>0,B>0,C>0$ should be feasible. – Cesareo Feb 08 '21 at 21:05
  • @Cesareo οκ so what you show us here is the steps of the method...to solve the problem...i understand – Nav89 Feb 08 '21 at 21:10
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    With the help of a symbolic processor we can obtain some more precise results. In MATHEMATICA try Reduce[{x a - nu g b - n g c >= 0, -nu g a + y b - n g c >= 0, -nu g a - nu g b + z c >= 0, x > 0, y > 0, z > 0, nu > 0, g > 0, n > 0, a > 0, b > 0,c > 0}, {a, b, c}] – Cesareo Feb 08 '21 at 21:32
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We can rewrite the system of inequalities, including the restrictions on $A,B,C$ to be positive, as follows:

$$ \begin{align*} A &\gt \frac{\nu g}{x} B + \frac{N g}{x} C \\ B &\gt \frac{\nu g}{y} A + \frac{N g}{y} C \\ C &\gt \frac{\nu g}{z} A + \frac{\nu g}{z} B \\ \\ A,B,C &\gt 0 \end{align*} $$

[Note that I've taken verbatim the terms as originally posted, but I asked for clarification about the third inequality in a Comment on the Question.]

The feasible region is convex, an intersection of six open half-spaces, each delineated by a plane passing through the origin. It is possible for this intersection to be empty, unlike the relaxed "weak inequalites" system where the origin $(0,0,0)$ will always be a solution.

Note that if $(A,B,C)$ is a feasible point, then so too is any positive scalar multiple $r\gt 0$ of it, $(rA,rB,rC)$. So geometrically, if the feasible region is not empty, it consists of a collection of open rays proceeding from the origin.

If the coefficients were known, we could represent the feasible region by taking its intersection with the plane:

$$ A + B + C = 1 $$

That intersection (if not empty) is an open polygon (the polygon excluding its perimeter). Such a polygon would be convex and have at most six sides, and its vertices can be readily computed by many linear programming packages. The feasible region of your problem is then all the open rays proceeding from the origin through a point in that open polygon. The points in the open polygon can be parameterized by barycentric coordinates.

If actual values of the coefficients can be given, I'd be happy to use them to illustrate this representation of solutions. Otherwise I could choose a few sets of coefficients to illustrate both the existence and non-existence of solutions.

hardmath
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