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.

• Where is $z$ in the formula? Also for the second inequality, it should be $x^2+y^2+2\geq(x+1)(y+1)$. Mar 8, 2021 at 20:03
• Sorry is a typo Mar 8, 2021 at 20:03
• Find the minimum of the function $x^2 + y^2 + 1 - (xy + x + y)$ and prove that it is non-negative. Mar 8, 2021 at 20:07
• Instead of going for $(x+1)(y+1)$ you can subtract $2xy$ to get $(x-y) ^2\geq (x-1)(1-y)$ or $(a+b) ^2\geq ab$ for $a=x-1,b=1-y$ Mar 9, 2021 at 0:35

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 \$$.

• Now you need to reverse the order of the implications. Mar 8, 2021 at 20:09
• Nice solution I never think in multiply by 2 Mar 8, 2021 at 20:18
• You are welcome! Mar 8, 2021 at 20:19
• Sometimes I don’t understand why some people don’t accept the proof that goes: “$x^2+y^2+1 \geq xy+x+y$ $\Leftarrow 2x^2+2y^2+2 - 2xy - 2x - 2y \geq 0$ $\Leftarrow (x-y)^2 + (x-1)^2 + (y-1)^2 \geq 0$, which holds for all $x, y \in \mathbb{R}$”. It’s logically correct, and saves you from having to copy the proof in reverse order again. Mar 8, 2021 at 20:19
• @BenjaminWang I am glad someone else has wondered about it too! I will remember that :) Mar 8, 2021 at 20:20

First Solution

By AM-GM inequality we have $$\frac{x^2+y^2}{2} \geq |xy| \geq xy \\ \frac{x^2+1}{2} \geq |x| \geq x \\ \frac{1+y^2}{2} \geq |y| \geq y \\$$

Second Solution: By Cauchy-Schwarz we have $$\left( xy+1\cdot x+y\cdot 1 \right)^2 \leq ( x^2+1^2+y^2)(y^2+x^2+1^2)$$

• Is really nice the first solution and the second I never think in use the Cauchy Schwarz inequality Mar 8, 2021 at 20:30
• IMHO, the factor two is too much, on the RHS, six times, in the AM-GM paragraph. Mar 28, 2021 at 18:28
• @Hanno ty fixed. Mar 29, 2021 at 3:19

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 $$x=\dfrac{y+1}{2}=\dfrac 22=1.$$

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}

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 $$f(x,y)\geq 0$$

• Do you mean convex instead of positive? Mar 8, 2021 at 20:30
• Yes, You are right
– user795628
Mar 8, 2021 at 20:35

Apply Cauchy-Schwarz or Buniakovsky inequality: $$(xy+x+y)^2 = (x\cdot y + 1\cdot x + y\cdot 1)^2 \le (x^2+1^2+y^2)(y^2+x^2+1^2) = (x^2+y^2+1)^2\implies xy+x+y \le x^2+y^2+1$$.

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 $$x < 1 and so $$(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)$$.

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 $$x=\dfrac{1+y}2=?$$

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