# Solve $\begin{cases} y<x\\ -(x-y)^2\leq 4 (1-y-x+xy) \end{cases}$

Solve the following system $$\begin{cases} y<x\\ -(x-y)^2\leq 4 (1-y-x+xy) \end{cases}$$ I tried to solve it graphically (e.g. $y=x$ is the bisector) but I can't handle the second inequality.

Maybe I should solve the second inequality algebraically in $y$ or $x$ and then set $y<x$.

Suggestions?

• Looking at it numerically, it seems like the second equation is always true (!) – Matti P. Jul 13 '18 at 9:25

You can rewrite the inequality as $$(x-y)^2+4xy -4(x+y)+4\ge 0\iff (x+y)^2 -4(x+y)+4\ge 0.$$
Hint* the second inequality is equivalent to $(x+y-2)^2>0$, so effectively it can be ignored in this case...
Hint. Note that the second inequality is always satisfied: $$-t^2+2ts-s^2=-(t-s)^2=−(x−y)^2\leq4(1−y−x+xy)=4ts\Leftrightarrow (t+s)^2\geq 0$$ where $t=x-1$ and $s=y-1$.