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!

Draw a diagram.  For $y\ge0$, the curve
$$y(y+1)=(x+1)^2$$
is the upper half of a hyperbola with turning point at $(-1,0)$.  One of the asymptotes of this hyperbola is $y=x+\frac{1}{2}$.  The inequality
$$y(y+1)\le (x+1)^2$$
defines the region below this hyperbola.  The hyperbola
$$y(y-1)=x^2$$
has turning point $(0,1)$ and is congruent to the first hyperbola; one of its asymptotes is also $y=x+\frac{1}{2}$; the second hyperbola is shifted upwards along the direction of this asymptote.  From a diagram it is therefore clear that the region lying below the first hyperbola also lies below the second.  Therefore, if the first inequality is true (and $y\ge0$) then so is the second.
