$\forall\ x,y,z\in \mathbb{R}$ 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!

We need to prove that
$$\left(|x+y+z|+|x|+|y|+|z|\right)^2\geq\left(|x+y|+|x+z|+|y+z|\right)^2$$ or
$$\sum_{cyc}\left(|x(x+y+z)|+|yz|\right)\geq\sum_{cyc}|(x+y)(x+z)|,$$
which is just a triangle inequality:
$$|x(x+y+z)|+|yz|\geq|x(x+y+z)+yz|=|(x+y)(x+z)|.$$
Also we can use Popoviciu.
For all convex function $f$ we have
$$f(x)+f(y)+f(z)+3f\left(\frac{x+y+z}{3}\right)\geq2\left(f\left(\frac{x+y}{2}\right)+f\left(\frac{x+z}{2}\right)+f\left(\frac{y+z}{2}\right)\right).$$
Since $f(x)=|x|$ is a convex function, we obtain
$$|x|+|y|+|z|+3\left|\frac{x+y+z}{3}\right|\geq2\left(\left|\frac{x+y}{2}\right|+\left|\frac{x+z}{2}\right|+\left|\frac{y+z}{2}\right|\right)$$ or
$$|x+y+z|+|x|+|y|+|z|\geq|x+y|+|x+z|+|y+z|$$
and we are done!
