Since
$$\det \begin{pmatrix} 1 & -x & -x \\ -y & 1 & -y \\ -z & -z & 1 \end{pmatrix} = 1 - 2xyz - xy - yz - zx,$$
the determinant of the system of equations
$$a - xb - xc = 0, \quad ya - b + yc = 0, \quad za + zb - c = 0, \quad (2)$$
in the variables $a, b, c$, is equal to zero by hypothesis. Hence, there are numbers $a, b, c$, not all of which are zero, that satisfy the system (2), in which case
$$a = (b+c)x, \quad b = (c+a)y, \quad c = (a+b)z.$$
Since $x, y, z$ are positive, none of $a, b, c$ is equal to zero and so $a(b+c), b(c+a)$ and $c(a+b)$ are also positive, whence $a, b, c$ have the same sign. The result follows.