Absolute!

Note that

$\begin{aligned} \dfrac{x-y+z}{2}+\dfrac{x+y-z}{2} & = x \\ \dfrac{-x+y+z}{2}+\dfrac{x+y-z}{2} & = y \\ \dfrac{x-y+z}{2}+\dfrac{-x+y+z}{2} & = z \end{aligned}$

Since $|a|+|b|\geq |a+b|$ ,

$\begin{aligned} \dfrac{|x-y+z|}{2}+\dfrac{|x+y-z|}{2} & \geq |x| \\ \dfrac{|-x+y+z|}{2}+\dfrac{|x+y-z|}{2} & \geq |y| \\ \dfrac{|x-y+z|}{2}+\dfrac{|-x+y+z|}{2} & \geq |z| \end{aligned}$

By adding the inequations, we get $\large |x|+|y|+|z|\leq |x-y+z|+|x+y-z|+|-x+y+z|$

Letting

$\begin{aligned} x & = a + b \\ y & = b + c \\ z & = c + a \end{aligned}$

then the following inequality is clear:

$\displaystyle |a + b| + |b + c| + |c + a| \leq 2|a| + 2|b| + 2|c|$