How can I prove $x^2+y^2-x-y-xy+1≥0$

Question

How can I prove

$$x^2+y^2-x-y-xy+1≥0$$

I tried $(x+y)^2-3xy-(x+y)+1≥0 \rightarrow(x+y-1)(x-y)-3xy+1≥0$ I can not continue

Answer 1

$$x^2+y^2-x-y-xy+1≥0 \Rightarrow \frac 12( 2x^2+2y^2-2x-2y-2xy+2)≥0 \Rightarrow \frac 12( x^2-2x+1+y^2-2y+1+x^2+y^2-2xy)≥0 \Rightarrow \frac 12((x-1)^2+(y-1)^2+(x-y)^2)≥0$$

Answer 2

Another way :$$\begin{align}x^2+y^2-x-y-xy+1&=x^2+(-1-y)x+y^2-y+1\\\\&=\left(x+\frac{-1-y}{2}\right)^2-\left(\frac{-1-y}{2}\right)^2+y^2-y+1\\\\&=\left(x-\frac{1+y}{2}\right)^2+\frac 34(y-1)^2\ge 0\end{align}$$

Answer 3

We need to prove that $$x^2-(y+1)x+y^2-y+1\geq0,$$ which is a quadratic inequality of $x$.

Thus, it's enough to prove that $$(y+1)^2-4(y^2-y+1)\leq0$$ or $$(y-1)^2\geq0,$$ which is obvious.

Answer 4

Let $u=x^2+y^2-x-y-xy\iff x^2-x(1+y)+y^2-y-u=0$

As $x$ is real, the discriminant must be $\ge0$

$$\implies(1+y)^2-4(y^2-y-u)\ge0$$

$$4u\ge3y^2-6y-1=3(y-1)^2-4\ge-4$$