Prove that $\forall x,y, \, \, x^2+y^2+1 \geq xy+x+y$ Prove that  $\forall x,y\in \mathbb{R}$ the inequality  $x^2+y^2+1 \geq xy+x+y$ holds.
Attempt
First attempt: I was trying see the geometric meaning, but I´m fall.
Second attempt: Consider the equivalent inequality given by $x^2+y^2\geq (x+1)(y+1)$
and then compare $\frac{x}{y}+\frac{y}{x} \geq 2 $ and the equality $(1+\frac{1}{x}) (1+\frac{1}{y})\leq 2$ unfortunelly not is true the last inequality and hence I can´t conclude our first inequality.
Third attempt:comparing  $x^2+y^2$ and $(\sqrt{x}+\sqrt{y})^2$ but unfortunelly I don´t get bound the term $2\sqrt{xy}$ with $xy$.
Any hint or advice of how I should think the problem was very useful.

We can rewrite $x^2+y^2+1 \geq xy+x+y \ $ as
$2x^2+2y^2+2 - 2xy - 2x - 2y \geq 0$
or as $(x-y)^2 + (x-1)^2 + (y-1)^2 \geq 0$
which holds for all $x, y \in \mathbb{R}$
Or start from $(x-y)^2 + (x-1)^2 + (y-1)^2 \geq 0$ and expand to show that $x^2+y^2+1 \geq xy+x+y \ $.
