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.

Answer
$$
\begin{align}
0
&\le\tfrac34((x-1)-(y-1))^2+\tfrac14((x-1)+(y-1))^2\\
&=(x-1)^2+(y-1)^2-(x-1)(y-1)\\
&=\left(x^2-2x+1\right)+\left(y^2-2y+1\right)-(xy-x-y+1)\\
&=\left(x^2+y^2+1\right)-(xy+x+y)
\end{align}
$$

Motivation
To prove $x^2+y^2+1\ge xy+x+y$, look at the difference
$$
\begin{align}
\left(x^2+y^2+1\right)-(xy+x+y)
&=(x-y)^2+(x-1)(y-1)\tag1\\
&=(u-v)^2+uv\tag2\\[2pt]
&=(u-v)^2+\tfrac14\left((u+v)^2-(u-v)^2\right)\tag3\\
&=\tfrac34(u-v)^2+\tfrac14(u+v)^2\tag4\\
&=\tfrac34((x-1)-(y-1))^2+\tfrac14((x-1)+(y-1))^2\tag5\\
&\ge0\tag6
\end{align}
$$
Explanation:
$(1)$: subtract $2xy+1$ from both sides of the minus sign
$(2)$: $u=x-1$ and $v=y-1$
$(3)$: $uv=\frac14\left((u+v)^2-(u-v)^2\right)$
$(4)$: combine terms
$(5)$: $u=x-1$ and $v=y-1$
$(6)$: sum of two non-negative numbers is non-negative
