How to prove $x^2 + y^2 + z^2\geq xy + xz + yz$

Question

Result: Let $, , ∈ ℝ$. Then we have $^2 + ^2 + ^2 ≥ + + $

Need some help proving this, just a few steps with work. Was thinking you start with $x^2+y^2+z^2−xy−xz−yz$ then factor? can anyone show me how to solve this?

$x^2+y^2+z^2≥xy+yz+zx⇔ $

$⇔2(x^2+y^2+z^2)≥2(xy+yz+zx)⇔$

$⇔x^2−2xy+y^2+y^2−2yz+z^2+z^2−2xz+x^2≥0⇔$

$⇔(x−y)^2+(y−z)^2+(z−x)^2≥0$

Should I do anything else? Or is this right?

Answer 1

$$^2 + ^2 + ^2 ≥ + + $$ $$2x^2+2y^2+2z^2 ≥ 2xy+2xz+2yz$$ $$(x^2-2xy+y^2)+(x^2-2xz+z^2)+(y^2-2yz+z^2) ≥ 0$$ $$(x-y)^2+(x-z)^2+ (y-z)^2≥0$$

Answer 2

Your proof is right, but maybe it looks a bit of better by using a cyclic sum: $$\sum_{cyc}(x^2-xy)=\frac{1}{2}\sum_{cyc}(2x^2-2xy)=\frac{1}{2}\sum_{cyc}(x^2-2xy+y^2)=\frac{1}{2}\sum_{cyc}(x-y)^2\geq0.$$

We can use also the following. $$x^2+y^2+z^2-xy-xz-yz=x^2-(y+z)x+y^2-yz+z^2=$$ $$=\left(x-\frac{y+z}{2}\right)^2+y^2-yz+z^2-\frac{(y+z)^2}{4}=\left(x-\frac{y+z}{2}\right)^2+\frac{3(y-z)^2}{4}\geq0.$$

Answer 3

Another way.

Let $y=x+u$ and $z=x+v$.

Thus, $$x^2+y^2+z^2-xy-xz-yz=$$ $$=x^2+(x+u)^2+(x+v)^2-x(x+u)-x(x+v)-(x+u)(x+v)=$$ $$=x^2+x^2+2xu+u^2+x^2+2xv+v^2-x^2-xu-x^2-xv-x^2-xv-xu-uv=$$ $$=u^2-uv+v^2=\left(u-\frac{v}{2}\right)^2+\frac{3v^2}{4}\geq0.$$