Inequality for absolute values

How do you show either of the equivalent inequalities:

$$2(|a|+|b|+|c|)\leq |a+b+c|+|a+b-c|+|a-b+c|+|a-b-c|$$

or $$|x+y|+|x+z|+|y+z|\leq |x|+|y|+|z|+|x+y+z|$$ Hold for complex numbers or in $n$ dimensions ?

See https://mathoverflow.net/questions/167685/absolute-value-inequality-for-complex-numbers where the same question is asked and the comments show that a proof must involve inner product properites of $\mathbb{C}$. Further the mathoverflow comments of Bill Johnson give a proof in great generality using sophisticated Banach space techniques. An elementary proof however would still be welcome.

Before this question was edited, a number of correct proofs for real values have been given using a case analysis technique, which, it must be remarked, form the base case for the arguments in mathoverflow.

Once the inequality is proved in ${\bf R}$, it follows in any inner-product space by writing each $|z|$ as a multiple of the average of $|u \cdot z|$ over unit vectors $u$. In ${\bf C}$ the formula is $$ |z| = \frac14 \int_0^{2\pi} \bigl| {\rm Re}(e^{i\theta} z) \bigr| \, d\theta. $$ Applying this to $z=a$, $b$, $c$, and $a \pm b \pm c$ reduces the desired inequality to the one-dimensional case. (Abridged from my answer in mathoverflow.)

I will consider a different form of:

$$|x+y|+|x+z|+|y+z|\leq |x|+|y|+|z|+|x+y+z|$$

Note that $$\begin{align}|x+y+z| &\leq |x+y| + |z| \leq |x| + |y| + |z|\\&\leq |x+z| + |y| \leq |x| + |y| + |z|\\&\leq |y+z| + |x| \leq |x| + |y| + |z|\end{align}$$

Summing the inequalities:

$$3|x+y+z| \leq |x+y|+|x+z|+|y+z| + |x| + |y| + |z| \leq 3|x| + 3|y| + 3|z|$$

We will first use: $$3|x+y+z| \leq 3|x| + 3|y| + 3|z| \qquad \qquad\ \ \ \ \ \ \ \ \ (1)$$

And then $$|x+y|+|x+z|+|y+z| + |x| + |y| + |z| \leq 3|x| + 3|y| + 3|z| \qquad\ \ \ \ \ \ \ \ \ (2)$$

Now I will prove that:

$$3|x+y|+3|x+z|+3|y+z|\leq 3|x|+3|y|+3|z|+3|x+y+z|$$

Using the first inequality $(1)$ we find that

$$3|x+y|+3|x+z|+3|y+z|\leq 6|x|+6|y|+6|z|$$

$$|x+y|+|x+z|+|y+z|\leq 2|x|+2|y|+2|z|$$

Add $|x| + |y| + |z|$ to both sides to find

$$|x+y|+|x+z|+|y+z| + |x| + |y| + |z| \leq 3|x| + 3|y| + 3|z|$$

Which is true from above $(2)$.