<s>[INST] $\forall\ x,y,z\in \mathbb{R}$ Show that: $|x+y|+|y+z|+|x+z|\leq |x+y+z|+|x|+|y|+|z|$ 
$\forall\ x,y,z\in \mathbb{R}$ Show that: $$|x+y|+|y+z|+|x+z|\leq |x+y+z|+|x|+|y|+|z|$$

i tired,
i notice that $x,y,z$ plays a symmetrical role in the inequality 
notice also that 
\begin{align*}
|x+y|+|y+z|+|x+z|\leq |x+y+z|+|x|+|y|+|z| & \Longleftrightarrow \\
 (|x+y|-|x|)+(|y+z|-|y|)+(|x+z|-|z|) \leq |x+y+z| 
\end{align*}
note that $\forall a,b\in \mathbb{R}\quad |a|-|b|\leq |a+b| $
then  $$(|x+y|-|x|)+(|y+z|-|y|)+(|x+z|-|z|) \leq |x|+|y|+|z|$$
i'm stuck here 
any help would be appreciated!
 [/INST] Let $a=x+y$, $b=x+z$ and $c=y+z$. Then, the inequality to be shown
can be rewritten as 
$$
2(|a|+|b|+|c|) \leq |a+b+c|+|a+b-c|+|a-b+c|+|-a+b+c| \tag{1}
$$
Let us put $f(a,b,c)=|a+b+c|+|a+b-c|+|a-b+c|+|-a+b+c|$. It is easy to
see that $f$ is even in each variable, so that $f(a,b,c)=f(|a|,|b|,|c|)$.
We may therefore assume that $a,b,c$ are all nonnegative, so that it suffices 
to show that
$$
2(a+b+c) \leq |a+b+c|+|a+b-c|+|a-b+c|+|-a+b+c|\tag{2}
$$
But (2) follows from the triangle inequality, since 
$$
2(a+b+c)=(a+b+c)+(a+b-c)+(a-b+c)+(-a+b+c) \tag{3}
$$
</s>