$\forall\ x,y,z\in \mathbb{R}$ 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!
 A: 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 
$$
2(a+b+c)=(a+b+c)+(a+b-c)+(a-b+c)+(-a+b+c) \tag{3}
$$
A: 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!
A: Prove that $\|a\|+\|b\| + \|c\| + \|a+b+c\| \geq \|a+b\| + \|b+c\| + \|c +a\|$ in the plane.
Note that indentity： \begin{align*}
&(|a|+|b|+|c|-|b+c|-|a+c|-|a+b|+|a+b+c|)(|a|+|b|+|c|+|a+b+c|)\\
&=(|b|+|c|-|b+c|)(|a|-|b+c|+|a+b+c|)+(|c|+|a|-|c+a|)(|b|-|c+a|+|a+b+c|)+(|a|+|b|-|a+b|)(|c|-|a+b|+|a+b+c|)
\end{align*}
