Austria2019

The inequality $(2x-y{)}^2\ge 0$ (with equality if and only if $y=2x$) is equivalent to $$(2x+y{)}^2\ge 8xy.$$ The constraint $(x+1)(y+2)=8$ gives $2x+y=6-xy$. Substituting this into the inequality above yields $$(6-xy{)}^2\ge 8xy,$$ which is equivalent to $$(xy-10{)}^2\ge 64.$$ As we noted already, equality holds for $y=2x$. In this case, the constraint becomes $(x+1)(2x+2)=8$ which yields $x=1$ or $x=-3$ and finally the two pairs $(x,y)=(1,2)$ and $(x,y)=(-3,-6)$. We easily verify that equality actually holds in both cases.