I was reading the question posted here:Does $y(y+1) \leq (x+1)^2$ imply $y(y-1) \leq x^2$?. The solution posted was:
"Given $y^2+y≤x^2+2x+1$, if possible, let $x2<y(y−1)$. 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 $y<x+1/2.$
Hence we also have $y−1<x−1/2.$ As $y>1$, the LHS is positive, and we can multiply the last two to get $y(y−1)<x^2−1/4⟹y(y−1)<x^2$, a contradiction." However, I don't think this is right because he had $y<x+1/2$, and in the last step he multiplied $y−1$ by $y$ and $x−1/2 by x+1/2$. However, this inequality is not in the same proportion anymore, because y does not equal x+1/2. It seems like he increased the value of the right side compared to the left, and then concluding that the right side is bigger/ It's like if I try to prove that $a \lt b$, and to solve this I say $a \lt b+5$. While this is true, this changes the inequality.