if $0\leq x < 1$ is obviously true if $x \ge 1$: $y^2 \geq x(x + 1)$<=> $y \ge \sqrt{x(x + 1)}$ We will prove: $\sqrt{x(x + 1)}-1 > \sqrt{x(x -1)}$ <=>$\sqrt{x(x + 1)}- \sqrt{x(x -1)}>1$ <=>$2x^2-2x\sqrt{(x+1)(x-1)}>1$ <=>$(2x^2-1)^2>4x^{2}(x+1)(x-1)$ <=>$1>0$(true).

Pinko wrote: $x \geq 0$ and $y$ are real numbers for which $y^2 \geq x(x + 1)$. Prove that: $(y - 1)^2 \geq x(x-1)$.

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