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

*

*Equality occurs if and only if, when $y=1$ and $x=\dfrac{y+1}{2}=\dfrac 22=1.$
A: Consider $f(x,y)=x^2+y^2+1-(xy+x+y)$ notice that
$\frac{\partial{f}}{\partial{x}}=2x-y+1$ and $\frac{\partial{f}}{\partial{y}}=2y-x+1$
then $\nabla f=0$ in $A=(-1,-1)$.
The function is convex since $f^{\prime \prime}(x,y)>0 $ hence the function have a minimum value in $1$ and then
$$f(x,y) \geq 1 >0 $$ and hence $$f(x,y)\geq 0$$
A: Just to give yet another approach, observe that if we let $x=u+v$ and $y=u-v$, then
$$\begin{align}
x^2+y^2+1\ge xy+x+y
&\iff(u+v)^2+(u-v)^2+1\ge(u+v)(u-v)+(u+v)+(u-v)\\
&\iff2u^2+2v^2+1\ge u^2-v^2+2u\\
&\iff u^2-2u+1+3v^2\ge0\\
&\iff(u-1)^2+3v^2\ge0
\end{align}$$
A: Apply Cauchy-Schwarz or Buniakovsky inequality:  $(xy+x+y)^2 = (x\cdot y + 1\cdot x + y\cdot 1)^2 \le (x^2+1^2+y^2)(y^2+x^2+1^2) = (x^2+y^2+1)^2\implies xy+x+y \le x^2+y^2+1$.
A: 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 $(y-x)^2 > (y-1)(1-x) = (x-1)(1-y)$.
A: 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
A: Let $z=x^2+y^2-xy-x-y$
$$\iff x^2-x(1+y)+y^2-y-z=0$$
As $x$ is real, the discriminant must be $\ge0$ of the above quadratic equation
i.e., $$(1+y)^2\ge4(y^2-y-z)$$
$$\iff4z\ge3y^2-6y-1=3(y-1)^2-4\ge-4$$
$$\implies z\ge-1$$ the equality occurs if $y=1$ and $x=\dfrac{1+y}2=?$
