<s>[INST] 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??
 [/INST] Let $c=x^2+y^2+1-(xy+x+y)$
$\iff x^2-x(1+y)+1-y+y^2-c=0$
As $x$ is real, the discriminant must be $\ge0$
i.e., $$(1+y)^2\ge4(1-y+y^2-c)\iff4c\ge3(1-y)^2$$  which is $\ge0$ for real $y$
Alternatively,
$$ x^2-x(1+y)+1-y+y^2=\left(x-\dfrac{1+y}2\right)^2+\dfrac{3(1-y)^2}4$$
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