A question on a proof of $y(y+1) \le (x+1)^2 \implies y(y-1) \le x^2$ 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.

To reword the proof given there:
We start with real numbers $x,y$ such that
$$\tag1 y\ge0 $$
$$\tag2 y^2+y\le x^2+2x+1$$
$$\tag3 x^2<y(y-1)$$
If we had $y\le1$, then the right hand side in $(3)$ would be the product of a nonnegative (according to $(1)$) and a nonpositve (by assumtion) factor, hence nonnegative, contradicting $x^2\ge0$. Therefore 
$$\tag4 y>1.$$
Adding $2x+1$ to $(3)$ and combining with $(2)$ we find
$y^2+y\le x^2+2x+1<y^2-y+2x+1$, i.e. 
$$\tag5 y\le x+\frac12.$$
By subtracting $1$, this becomes
$$\tag6 y-1\le x-\frac12.$$
Because of $(4)$ and $(6)$, the number $x-\frac12$ is positive, hence we are allowed to multiply $(5)$ with $x-\frac12$. Also, we can multiply $(6)$ with the (again by $(4)$) positive number $y$ and combine this to find
$$ x^2\stackrel{(3)}<y(y-1)\stackrel{y\cdot(6)}\le y\left(x-\frac12\right)\stackrel{(x-\frac12)\cdot(5)}\le \left(x+\frac12\right)\left(x-\frac12\right)=x^2-\frac14<x^2,$$
contradiction.
