TITLE: Proving $|x|+|y|+|z| ≤ |x+y-z|+|y+z-x|+|z+x-y|$ for all real $x$
QUESTION [5 upvotes]: We need to prove that for all real $x,y,z$
$$|x|+|y|+|z| ≤ |x+y-z|+|y+z-x|+|z+x-y|$$
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I don't know how to solve in mod form so I thought about squaring and removing the mod, but the mod on RHS still persists.
I believe there can be a different approach to it, I would like to have some hints on how to do approach it.

REPLY [3 votes]: 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}