# Prove that for $\forall x,y,z \in \mathbb{R}, x^2+y^2+z^2\geq xy+yz+xz$

Show that for $\forall x,y,z \in \mathbb{R},x^2+y^2+z^2\geq xy+yz+xz$.

I first assumed that $x\geq y \geq z$, but I'm having problems with the $z^2$. By itself, $z^2$ is clearly not greater than any of products of the other terms so I tried to show that $z^2+y^2$ was greater than something but failed. Can someone explain how I could prove this?

We have $(x-y)^2+(y-z)^2+(z-x)^2\ge0$. Expand, and we will get our inequality. And as a bonus we get that there is equality if and only if $x=y=z$.

All we need is the trivial inequality $x^2 \geq 0$ for all $x$. Notice

$$2(x^2 + y^2 + z^2 - xy - yz - zx) = x^2 + x^2 + y^2+y^2 + z^2 + z^2 - 2xy - 2yx - 2xz = (x-y)^2 + (y-z)^2 + (z-x)^2 \geq 0$$

Hence,

$$2(x^2 + y^2 + z^2 - xy - yz - zx) \geq 0 \implies x^2 + y^2 + z^2 - xy - yz - zx \geq 0 \implies x^2 + y^2 + z^2 \geq xy + yz +zx$$

Another way to solve this problem. Using Cauchy-trick,

$$(ab+bc+ca)^2 \leq (a^2 + b^2 + c^2)(b^2 + c^2+a^2) = (a^2+b^2+c^2)^2$$

Squaring, gives desired result.

$$x^2 + y^2 + z^2 - yz - zx - xy = \frac{3}{4} (-x+y)^2 + \frac{1}{4} (-x-y + 2 z)^2$$