I have the following system of linear inequality on $x_1, x_2, \dots, x_n$, $x_i \in \mathbb{R} \; \forall i$

$x_i - 2x_j < b \; \forall i, j$ The right hand side of the inequality ($b \in \mathbb{R}_+$) is fixed (and given) for all inequalities.

From simulation I can see that the only solution to this system of equation is when all $x_i$'s are equal. Is there a formal proof to show this ?

  • $\begingroup$ I'm confused. If say $b=4$ and $x_i=0,1$. then that's a perfectly good solution. $\endgroup$
    – Alex R.
    Jul 13, 2015 at 19:44
  • $\begingroup$ Basically it sums up to solving $x_i-2x_j<0$ if $b$ is the same for all the equations and since it is strictly positive. Maybe we are missing something, but an idea would be to show by contradiction that if a pair of those is not equal, then you may choose $x_i=3x_j$ for some $i,j$, which will invalidate the RHS. *Note: basically construct some $x$ that will provide a counterexample. $\endgroup$ Jul 13, 2015 at 19:57

1 Answer 1


You need to have some restriction on $b$.

Otherwise, for any set of $x_i$, choose $b > 3\max(|x_i|) $. Then all the inequalities hold.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .