Suppose $x \ge 0$ and $y^2 \ge x(x+1)$. If $0 \le x \le 1$, then $(y-1)^2 \ge 0 \ge x(x-1)$. If $x > 1$, either $y \ge \sqrt{x(x+1)}$ or $y \le -\sqrt{x(x+1)}$. If $y \ge \sqrt{x(x+1)}$, then $y > 1$, since $x > 1$. So $(y-1)^2 \ge (\sqrt{x(x+1)}-1)^2 = x^2+x+1-2\sqrt{x(x+1)}$. It suffices to prove that this is $\ge x(x-1)$. That is, $2x+1 \ge 2\sqrt{x(x+1)}$, when $x > 1$. Squaring both sides, this reduces to $1 > 0$, which is true.
The second case is $y \le -\sqrt{x(x+1)}$. In this case, we have $(y-1)^2 \ge (\sqrt{x(x+1)}+1)^2 = x^2+x+1+2\sqrt{x(x+1)} > x^2+x > x^2-x$, which proves the result.