I am given $3$ points $(x\_i,y\_i,z\_i) \in \mathbb{R}^3\setminus \{\mathbf{0} \}$, for $i=1,2,3$, satisfying the following two polynomial equations (the first equation is actually not the intended one, see edit 1):


$$3+z\_1+z\_2+z\_3-\frac14(z\_1x\_2x\_3 + z\_2x\_3x\_1 + z\_3x\_1x\_2) - \frac14(z\_1y\_2y\_3 + z\_2y\_3y\_1 + z\_3y\_1y\_2) - \frac12(x\_1x\_2+x\_2x\_3+x\_3x\_1) - \frac12(y\_1y\_2+y\_2y\_3+y\_3y\_1) = 0$$


and


$$z\_1(x\_2y\_3-x\_3y\_2) + z\_2(x\_3y\_1-x\_1y\_3) + z\_3(x\_1y\_2-x\_2y\_1) = 0$$


Notice that the two polynomial equations are invariant when cyclically permuting the $3$ points in $\mathbb{R}^3$.


Prove that there does not exist $c\_i > 0$, such that


$$\mathbf{0} = \sum\_{i=1}^3 c\_i \mathbf{x}\_i$$


where $\mathbf{x}\_i = (x\_i,y\_i,z\_i)$ ($1 \leq i \leq 3$). Any ideas, anyone?


edit 1: there is a mistake in the first equation. I apologize. If we let $r\_i = \| \mathbf{x}\_i \|$, for $i=1,2,3$, then the first equation should read instead:


$r\_1(r\_2+z\_2)(r\_3-z\_3)-r\_1(x\_2x\_3+y\_2y\_3) + \text{ cyclic } = 0$.


where cyclic denotes cyclically permuting $1$, $2$ and $3$. Thus the first equation amounts simply to:


$r\_1(r\_2r\_3-(\mathbf{x}\_2,\mathbf{x}\_3)) + \text{ cyclic } = 0$


where $(-,-)$ denotes the Euclidean inner product in $\mathbb{R}^3$. Notice that the second equation is just that the $\mathbf{x}\_i$ are linearly independent over $\mathbb{R}$, and thus lie in a two-dimensional subspace of $\mathbb{R}^3$. Back to the first equation, we can now apply Cauchy-Schwarz (the equality part of Cauchy-Schwarz), to get that the $x\_i$ are actually all in the same direction, so that they are all positive multiples of each other. Then one can deduce that $\mathbf{0}$ cannot be in the convex hull of the $\mathbf{x}\_i$. This is the answer actually. 


Thank you Iosef Pinelis for your answer, and I apologize, because I had the first equation wrong at the beginning, and you had kindly provided a counterexample based on my wrong first equation! Sorry about that.



How to show that the origin is not in the convex hull, in this problem?
I actually answered my question in edit 1. I apologize to Iosif Pinelis who had found a counterexample to my original version of the question, but after correcting my question (which I got via a calculation), the answer follows easily from Cauchy-Schwarz (see edit 1). This post is probably confusing to readers, but I hope edit 1, which contains the answer, is clear enough.


