prove that $x^2 + y^2 + z^2 \ge xy + xz + yz$
I tried proving it the way I prove $x^2 + y^2 \ge 2xy$. But it just doesn't seem to work. I tried using AM-GM. Should I be using GM-HM or AM-HM? I really don't know what the problem is and I'd appreciate any help and a full proof explanation. Thanks.
The inequality $x^2+y^2 \geq 2xy$ works here. \begin{align*} x^2+y^2 \geq 2xy \\ y^2+z^2 \geq 2yz\\ z^2+x^2 \geq 2zx \end{align*} Adding those 3 inequalities gives $$ x^2+y^2 + y^2+z^2 +z^2+x^2= 2(x^2+y^2+z^2)\geq 2(xy+yz+zx)$$ So canceling 2 from both sides gives you the inequality.
$$(x - y)^2 + (y - z)^2 + (z - x)^2 \geq 0$$
hence
$$2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx \geq 0 $$ $$x^2 + y^2 + z^2 \geq xy + yz + zx$$
We see that for all real numbers $x$, $y$, and $z$ that $(x - y)² + (x - z)² + (y - z)² ≥ 0$ by virtue of being a sum of squares. Expanding the left hand side gives $2x² + 2y² + 2z² - 2xy - 2xz - 2yz ≥ 0$. Dividing through by two and adding $xy + xz + yz$ to both sides of the inequality gives the desired result $x² + y² + z² ≥ xy + xz + yz.$
Using the AM-GM inequality: $$ xy + xz + yz \leq \\ \sqrt{x^2y^2} + \sqrt{x^2z^2} + \sqrt{y^2 z^2} \leq\\ \frac{x^2 + y^2}{2} + \frac{y^2 + z^2}{2} + \frac{x^2 + z^2}{2} =\\ x^2 + y^2 + z^2 $$
Suppose $x:={\rm mid}\{x, y, z\}$, then
$$\sum x^{2}- \sum xy= (y+ z- 2x)^{2}- 3(x- y)(x- z)\geq 0$$