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.

If you look at the inequality as a quadratic inequality with respect to the variable $x$, then we have
$$x^2+y^2+1 \geq xy+x+y$$
$$\implies x^2+y^2+1 -xy-x-y≥0$$
$$\implies x^2-x(y+1)+(y^2-y+1)≥0$$
$$\implies \left( x -\frac{y+1}{2}\right)^2-\left(\frac{y+1}{2}\right)^2+y^2-y+1≥0$$
$$\implies \left( x -\frac{y+1}{2}\right)^2+ \frac 34 (y - 1)^2≥0.$$

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*Equality occurs if and only if, when $y=1$ and $x=\dfrac{y+1}{2}=\dfrac 22=1.$
