CatalinBordea wrote: Find all real numbers $ a $ such that $ x,y>a\implies x+y+xy>a. $ If $a<-1$, choosing $(x,y)=(1,\frac{a-1}2)$ gives contradiction. If $a=-1$, condition is $(x+1)(y+1)>0$ $\forall x,y>-1$, true and such $a$ fits If $0>a>-1$, choosing $x=y=\sqrt{a+1}-1$ gives contradiction. If $a\ge 0$ $x,y>a\ge 0$ implies $x+y+xy\ge x>a$, and so all such $a$ fit Hence the answer $\boxed{a\in\{-1\}\cup[0,+\infty)}$