# Putnam Exam question [duplicate]

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!

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## marked as duplicate by Aryabhata, vadim123, Hurkyl, Live Forever, HakimJul 14 '14 at 1:46

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.
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 $y > 1$.