Putnam Exam question Prove or disprove: if $x$ and $y$ are real numbers with $y\ge 0$ and $y(y+1)\le (x+1)^2$, then $y(y-1)\le x^2$.
How should I approach this proof?  The solution starts with assuming $y\ge 0$ and $y\le 1$, but I'm not sure how to arrive at that second assumption or go from there.  Thank you in advance!

If $0 \le y \le 1$ the statement is trivially true, because $y(y-1) \le 0 \le x^2$.  So now you need to see what happens when $y > 1$.
