You can show this directly.
Multiplying the constraints gives $x_1 x_2 \ge x_2(x_2+2)$. Since
$\min_x x(x+2) = -1$, we have
$x_1 x_2 \ge -1$. Furthermore, since $x(x+2) = -1$ iff $x=-1$, we see that,
in fact, $x_1 x_2 > -1$ for $x_2 \neq -1$.
Evaluating the cost at $(1,-1)$ gives $-1$, so we see that $(1,-1)$ is a global minimiser. We have already shown that this is strict when $x_2 \neq -1$, so now consider $x_2 = -1$ in which case the first constraint gives $x_1 \le 1$. If
$x_1 < 1$, then $x_1 x_2 = - x_1 > -1$, from which it follows that
$(1,-1)$ is a strict global minimiser.