Prove completing the square

Question

Prove that: $x + y + xy - x^2 - y^2 \leq 1$

If I use $-x^2 + xy - y^2$ to start completing the square, I get: $x + y -((x+y)(x-y)) - xy$

I am confused on how to keep going.

Answer 1

$\begin{align} f(x, y) &=x+y+xy-x^2-y^2\\ &=x+y-xy+2xy-x^2-y^2\\ &=x+y-xy-(x-y)^2\\ &=1-1+x+y-xy-(x-y)^2\\ &=1-(x-1)(y-1)-(x-y)^2\\ \end{align} $

If we can show that $g(x, y) = (x-y)^2 + (x-1)(y-1) \ge 0$, we are done.

$\begin{align} g(x+1, y+1) &=(x-y)^2 + xy\\ &=x^2-xy+y^2\\ &=x^2-xy+y^2/4 + 3y^2/4\\ &=(x-y/2)^2+3y^2/4\\ &\ge 0\\ \end{align} $

so we are done.

Answer 2

$$x + y + xy -x^2 -y^2 = -\frac{1}{4} \left((x + y - 2)^2 + 3(x-y)^2\right) + 1 \leq 1$$