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.

Look for perfect squares
$x^2 + y^2 + 1 \ge xy + x+y\iff $
$x^2 - 2xy+y^2 \ge -xy + x + y - 1\iff$
$(x-y)^2 \ge x+y - xy - 1$.
Now as the LHS is $\ge 0$ if we can show the RHS is $\le 0$ that be great.  If not... well we may hit an inspiration on the way.  We can always factor
$x+y - xy - 1 = x-xy + y-1 = x(1-y) + y-1) = x(1-y)-(1-y) = (x-1)(1-y)$.
Hmmm, no need for that to be negative but if it is positive then either $(x-1)$ and $1-y$ are both positive or both negative.
If both are positive then $x > 1$ and $y< 1$.  So $x-y > x-1$ and $x-y > 1-y$ and so $(x-y)(x-y) > (x-1)(1-y)$.
And if both are negative then $x < 1<y$ and so $(x-y)^2 = (y-x)^2$ and $y-x > y-1>0$ and $y-x > 1-x>0$ and $(y-x)^2 > (y-1)(1-x) = (x-1)(1-y)$.
