Given any 12 real constants $a_1,a_2,a_3,b_1,b_2,b_3,c_1,c_2,c_3,d_1,d_2,d_3$ all $>0$ I'm asking myself if is it always possible to find positive reals $x,y,z,t,u,v$ such that


But I can not find an answer since I don't know how to work with the condition of $x,y,z,t,u,v$ positive.

What do you think? If the answer is "no" is there a condition on $a_1,a_2,a_3,b_1,b_2,b_3,c_1,c_2,c_3,d_1,d_2,d_3$ to guarantee that there exist such $x,y,z,t,u,v$?

As pointed out by user Marty Cohen one could substitute the variables $x,y,\dots$ with the variables $x^2,y^2,\dots$ and get quadratic equation. At this point one could introduce another variable $k\in \mathbb{R}$ and ask if the following system

$$a_1x^2+a_2y^2+a_3z^2=k\\ b_1t^2+b_2u^2+b_3x^2=k\\ c_1t^2+c_2y^2+c_3v^2=k\\ d_1v^2+d_2u^2+d_3z^2=k$$

has a solution such that all $x,y,z,t,u,v$ are $\neq 0$. Note that this is the intersection of 4 quadrics of $\mathbb{R}^6$

Edit: I'm searching conditions on the $a_1,a_2,a_3,b_1,b_2,b_3,c_1,c_2,c_3,d_1,d_2,d_3$ which guarantee the existence of $x,y,zt,u,v>0$, I don't want to solve the system with a linear program

  • $\begingroup$ Set $a_1=a_2=a_3=0$ and $b_1,b_2,b_3$ all positive. No matter what you choose for $x,y,z,t,u,v$ the first sum will be zero and and the second sum will be positive. $\endgroup$ Jun 1, 2017 at 12:43
  • $\begingroup$ the constants $a_1,a_2,a_3,b_1,b_2,b_3,c_1,c_2,c_3,d_1,d_2,d_3$ are given and all $\neq 0$ $\endgroup$
    – user00169
    Jun 1, 2017 at 12:44
  • $\begingroup$ Ok, then take $a_1=a_2=a_3=1$ (making the first sum positive) and $b_1=b_2=b_3=-1$ (making the second sum negative). $\endgroup$ Jun 1, 2017 at 12:59

2 Answers 2


To force them to be positive, use $x^2, y^2, ...$ instead of $x, y, ...$

  • $\begingroup$ Thank you, if I understand correctly you're saying that we can search for reals $x,y,z,t,u,v$ such that $a_1x^2+a_2y^2+a_3z^2=b_1t^2+b_2u^2+b_3x^2=c_1t^2+c_2y^2+c_3v^2=d_1v^2+d_2u^2+d_3z^2$ but I still can't solve it.. $\endgroup$
    – user00169
    Jun 1, 2017 at 12:34
  • $\begingroup$ Probably makes it harder since it's nonlinear. $\endgroup$ Jun 1, 2017 at 12:47
  • $\begingroup$ and the solution must be such that all $x,y,z,t,u,v$ are $\neq 0$ $\endgroup$
    – user00169
    Jun 1, 2017 at 12:50

For given positive real constants $$a_1,a_2,a_3,b_1,b_2,b_3,c_1,c_2,c_3,d_1,d_2,d_3$$ the question as to whether there are positive real numbers $x,y,z,t,u,v$ satisfying the system of equations $$a_1x+a_2y+a_3z=b_1t+b_2u+b_3x=c_1t+c_2y+c_3v=d_1v+d_2u+d_3z$$ is equivalent to the question of the feasibility of the linear program \begin{align*} a_1x+a_2y+a_3z&=p\\[4pt] b_1t+b_2u+b_3x&=p\\[4pt] c_1t+c_2y+c_3v&=p\\[4pt] d_1v+d_2u+d_3z&=p\\[4pt] x,y,z,t,u,v,p &\ge 1\\[4pt] \end{align*} However the above linear program is not always feasible.

For example, there's no feasible solution when \begin{align*} a_1,a_2,a_3 &= 2,1,3\\[4pt] b_1,b_2,b_3 &= 2,3,1\\[4pt] c_1,c_2,c_3 &= 3,3,3\\[4pt] d_1,d_2,d_3 &= 1,1,1\\[4pt] \end{align*} Update:

My answer was posted in response to your original question, before your latest edit. In your edited version, you state that you want to avoid using a linear program.

But it really is a linear program.

As to whether the special form of your equations allows for a general solution without methods equivalent to linear programming, it's possible, but I doubt it.


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