Proving $|x|+|y|+|z| ≤ |x+y-z|+|y+z-x|+|z+x-y|$ for all real $x$

Hint:

$$2x=(x+y-z)+(x+z-y) \implies x=\frac{x+y-z}{2}+\frac{x+z-y}{2}$$ Now repeat the process for $y$ and $z$ and use the triangle inequality trice

write $A=x+y-z, B=y+z-x, C=x+z-y$. Then $A+B+C=x+y+z$ which is not so useful. Instead use $$A+B = 2y,\\B+C=2z,\\C+A=2x$$ and so \begin{align} |x|+|y|+|z| &= \frac{|A+B|+|B+C|+|C+A|}2 \\ &\le \frac{|A|+|B| + |B| + |C| + |C| + |A|}2 \\ &= |A|+|B|+|C| \\ &=|x+y-z|+|y+z-x|+ |x+z-y|. \end{align}