Ireland

Since $$\det \left(\begin{array}{ccc}1& -x& -x\\ -y& 1& -y\\ -z& -z& 1\end{array}\right)=1-2xyz-xy-yz-zx,$$ the determinant of the system of equations $$a-xb-xc=0,ya-b+yc=0,za+zb-c=0,(2)$$ 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 $$a=(b+c)x,b=(c+a)y,c=(a+b)z.$$ 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.