Does $y(y+1) \leq (x+1)^2$ imply $y(y-1) \leq x^2$? Can anyone see how to prove the following?
If $x$ and $y$ are real numbers with $y\geq 0$ and $y(y+1) \leq (x+1)^2$ then $y(y-1) \leq x^2$.
It seems it is true at least according to Mathematica.
 A: Given $y^2 + y \le x^2 + 2x + 1$,  if possible, let $x^2 < y(y-1)$.  Clearly $y > 1$.  
Then $x^2 + (2x + 1) < y^2 - y + (2x + 1)$
So $y^2 + y < y^2 - y + 2x + 1$, which resolves to $y < x + \frac{1}{2}$. 
Hence we also have $y - 1 < x - \frac{1}{2}$.
As $y > 1$, the LHS is positive, and we can multiply the last two to get
$y(y-1) < x^2 - \frac{1}{4} \implies y(y-1) < x^2$, a contradiction.
A: Yes, it's true.
First note that if $0 \le y \le 1$, then $y - 1 \le 0$, so $y(y-1) \le 0 \le x^2$, and the inequality is true.
So we can assume $y > 1$. Then, as $y(y+1) = (y + \frac12)^2 - \frac14$ (by completing the square), the given $y(y+1) \le (x+1)^2$ becomes $(y + \frac12)^2 - \frac14 \le (x+1)^2$, and in particular $(y + \frac12)^2 \le (x+1)^2$ so $y \le x+\frac12$.
Then $y(y-1) = (y-\frac12)^2 - \frac14 \le x^2 - \frac14$, which is a stronger inequality than what you wanted.
If you're interested in how to arrive at this, the first version of this answer does this whole thing in a more straightforward (but longer) way; this is cleaner.
