Answer
$y(y+1)≤(x+1)^2$ is equal to $ y(y−1)≤x^2$
Work Step by Step
Let: y= $y-1$ and x= $x-1$
1. Substitute $y-1$ and $x-1$ in the given equation:
$y(y+1) \leq (x+1)^{2}$ = $y-1(y-1+1) \leq (x-1+1)^{2}$
2. Simplify
$(y-1)y \leq x^{2}$
3. Therefore:
$y(y+1)≤(x+1)^2 = y(y−1)≤x^2$
