Rewrite the inequality $x^2 + y^2 + 1 \ge 2(xy - x + y)$ as $x^2 - 2xy + y^2 + 2x - 2y + 1 \ge 0$. If we further rearrange the left-hand side into $(x - y)^2 + 2(x - y) + 1 \ge 0$, we notice that it is a perfect square, and the inequality becomes $((x - y) + 1)^2 \ge 0$. Hence, the inequality holds for all real $x$ and $y$. The equality case occurs if and only if $(x - y) + 1 = 0$.