1. Notice that $(x+y-z)+\left(z_{c}+x-y\right)=2 x$, so, $x+y-z+z+x-y \mid \geqslant$ 2| $x$ |, similarly, $|y+z-x|+|z+x-y| \geqslant 2| z|| x+y-,z|+| y+z-x \mid$ $2|y|$ , adding these three inequalities and dividing by 2 yields.