How to show that the origin is not in the convex hull, in this problem?

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.

• Perhaps you can apply Grobner bases? What makes it only work with cyclic permutations? Maybe one can reduce it a bit by introducing something similar to elementary symmetric polynomials.. – Per Alexandersson Apr 8 '18 at 18:51

A counterexample is $c_1= 2,c_2= 2,c_3= 81,x_1= -8,x_2= -1,x_3= \frac{2}{9},y_1= 0,y_2= 0,y_3= 0$, $z_1= 0,z_2= 0,z_3= 0$.
Added in response to the OP's comment: With the additional condition that the points $\mathbf x_i=(x_i,y_i,z_i)$ be nonzero, we still have a counterexample: $c_1= 1,c_2= 1,c_3= \frac{5233}{128},x_1= -8,x_2= -1,x_3= \frac{1152}{5233},y_1= \frac{7}{8},y_2= 0,y_3= -\frac{112}{5233}$, $z_1= 0,z_2= 0,z_3= 0$.
• oops I forgot to mention the $3$ points are in $\mathbb{R}^3 \setminus 0$. None of them can be at the origin. – Malkoun Apr 8 '18 at 20:13
• I have added a counterexample satisfying also the additional condition that the points $\mathbf x_i=(x_i,y_i,z_i)$ be nonzero. – Iosif Pinelis Apr 8 '18 at 20:26