I have the system of these three equations:
$$ax = y+z$$ $$by = x+z$$ $$cz = x+y$$
How do I find all $a$, $b$ and $c$ for which the system has real, positive solutions for $x$, $y$ and $z$?
As a comparison, I have a simpler system:
$$ax = y$$ $$by = x$$
For this simpler system, with a substitution I get that $ab = 1$ yields a system that has real positive solutions. I'd like to find something similar for the above system as well. Basically, I'm looking for the relation between $a$, $b$ and $c$, independent from $x$, $y$ and $z$.
This is somewhat similar to this other question, only my equations are different.
Multiply the second equation by $c$ to obtain $$bcy=cx+cz=cx+x+y$$ using the third equation, so that $$(bc-1)y=(c+1)x$$
Likewise multiply the first equation by $c$ to obtain equivalently $$(ac-1)x=(c+1)y$$Now multiply this second equation by $c+1$ so that $$(c+1)^2y=(ac-1)(c+1)x=(ac-1)(bc-1)y$$whence $y=0$ or $$c^2+2c+1=abc^2-ac-bc+1$$ so $c=0$ or $$abc-a-b-c=2$$
Note that the condition $c=0$ can be avoided (by symmetric argument) unless $a=b=c=0$
Further note that positive solutions imply $a,b,c\gt 0$ - because e.g. $a=\frac {y+z}x\gt0$.
Also the solution $a=b=c=-1$ has us multiplying by $c+1=0$ which is inadmissible if we want to remain strictly positive.
Suppose we have positive values $a,b,c,x$ - we then find $$y=\frac {ac-1}{c+1}x; z=\frac{ab-1}{b+1}x$$which are positive provided $ac,ab\gt 1$
Writing the condition $abc-a-b-c=2$ in the form $$b(ac-1)=2+a+c$$ the right-hand side is positive, $b$ is positive, so $ac\gt 1$ is enforced by the condition.
So if $a,,b,c\gt 0$ satisfy the condition, there is a unique positive solution for any chosen positive value of $x$.