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.

First Solution
By AM-GM inequality we have
$$
\frac{x^2+y^2}{2} \geq |xy| \geq xy \\
\frac{x^2+1}{2} \geq |x| \geq x \\
\frac{1+y^2}{2} \geq |y| \geq y \\
$$
Add them together.
Second Solution: By Cauchy-Schwarz we have
$$
\left( xy+1\cdot x+y\cdot 1 \right)^2 \leq ( x^2+1^2+y^2)(y^2+x^2+1^2)
$$
