Find non-negative solutions of systems of equations: $$\begin{cases} x^2y^2+1=x^2+xy \\ y^2z^2+1=y^2+yz \\ z^2x^2+1=z^2+zx \end{cases} $$
My work so far:
1) $(1;1;1) - $ solution.
2) $(y^2-1)x^2-yx+1=0$
$D=y^2-4(y^2-1)=-3y^2+4\ge0 \Rightarrow 0\le y \le \frac2{\sqrt3}$
3) $x^2y^2+1-(y^2z^2+1)=x^2+xy-(y^2+yz)$
$y^2(x-z)(x+z)=(x-y)(x+y)+y(x-z)$
Put $y=hx$. Then the first equation gives $h^2x^4-(h+1)x^2+1=0$. Hence $(h+1)^2\ge (2h)^2$, so $h\le 1$. Similarly, putting $z=ky$ we get $k\le 1$ from the second equation. But then $x=\frac{1}{hk}z$ and the third equation gives $hk\ge1$, so we must have $x=y=z$. Now the first equation gives $x^4-2x^2+1=0$ so $x^2=1$ and hence $x=1$. So the unique solution is $x=y=z=1$.