to prove $x^2 + y^2+1\ge xy + y + x$ $$x^2 + y^2+1\ge xy + y + x$$
$x$ and $ y$ belong to all real numbers
my attempt
$(u-2)^2\ge0\Rightarrow \frac{u^2}{4}+1\ge u $
let $u=x+y\Rightarrow \frac{(x+y)^2}{4}+1\ge x+y$
$\Rightarrow (x+y)^2+1\ge \frac{3}{4}(x+y)^2+(x+y)$
$but \frac{(x+y)^2}{4} \ge xy $ by AM-GM inequality
$\Rightarrow (x+y)^2+1\ge \frac{3}{4}(x+y)^2+(x+y)\ge3xy+(x+y)$
hence  $\Rightarrow x^2 + y^2+2xy+1\ge 3xy+x+y$
are the steps correct and is there any other better way??

Prove
$$f(x,y)=x^2+y^2+1-x-xy-y\geq 0.$$
$$f_x=2x-1-y$$
$$f_y=2y-1-x$$
$$f_x=0\implies y=2x-1$$
$$f_y=0\implies x=2y-1$$
$$y=2(2y-1)-1=4y-3\implies y=1$$
$$x=2(2x-1)-1=4x-3\implies x=1$$
There's a stationary point at $(1,1)$.
$$f_{xx}=2,f_{xy}=f_{yx}=-1,f_{yy}=2.$$
Since $$f_{xx}f_{yy}-f_{xy}^2=4-(-1)^2=3>0\text{ and }f_{xx}=2>0$$
then that stationary point is a minimum. Furthermore $f(1,1)=0\geq 0$.
