$\left(x+y+z\right)\left(x+y-z\right)\le 0\left(1\right)$
$\left(x+y+z\right)\left(x-y+z\right)\le 0\left(2\right)$  
$\left(x+y+z\right)\left(-x+y+z\right)\le 0\left(3\right)$

Dodajemy stronami nierówności $\left(1\right)$, $\left(2\right)$ i $\left(3\right)$:
$\left(x+y+z\right)\left(x+y-z\right)+\left(x+y+z\right)\left(x-y+z\right)+\left(x+y+z\right)\left(-x+y+z\right)\le 0$
$\left(x+y+z\right)\left(x+y-z+x-y+z-x+y+z\right)\le 0$
$\left(x+y+z\right)\left(x+y+z\right)\le 0$
${\left(x+y+z\right)}^2\le 0$
$x+y+z=0$          $■$

Przykład 9. Rozwiąż nierówność 

$\left|x-1\right|+\left|x+1\right|>\left|x\right|$

$\left|x-1\right|+\left|x+1\right|\ge \left|x-1+x+1\right|=\left|2x\right|=2\left|x\right|\underset{x\ne 0}{>}\left|x\right|$
Dla $x=0$:
$\left|0-1\right|+\left|0+1\right|=2>\left|0\right|$

$x\in \mathbb{R}$