Inequality for absolute values

Question

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.

Answer 1

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 $(2)$.

Answer 2

Case 1: $x + y + z = 0$, then both sides equal and we have equality.

Case 2: $x + y + z \neq 0$, then we may assume that $x + y + z = 1$ and the $2$ nd inequality is equivalent to: $|1-z| + |1-x| + |1-y| \leq 1 + |x| + |y| + |z|$

in this case, there are subcases:

a) $x \leq 1, y \leq 1, z \leq 1$: then $LHS = 1 - x + 1 - y + 1 - z = 3 -(x+y+z) = 3 - 1 = 2 = 1 + 1 = 1 + |x+y+z| \leq 1 + |x| + |y| + |z| = RHS$.

b) $x > 1$, $y \leq 1$, $z \leq 1$:then $LHS = 1 - z + x - 1 + 1 - y = x + (1 - y - z) = x + x = 2x = 2|x| = |x| + |1-y-z| \leq 1 + |x| + |y| + |z| = RHS$

c) $x > 1$, $y > 1$, $z \leq 1$:then $LHS = x - 1 + y - 1 + 1 - z = x + y - z - 1 \leq |x + y - z - 1| \leq |x| + |y| + |-z| + |-1| = 1+ |x| + |y| + |z| = RHS$

Answer 3

Both sides of the inequality are invariant under the sign change of $a$, $b$, or $c$. Furthermore, they are both symmetric with respect to $a$, $b$, and $c$. Therefore, without loss of generality we can assume that $0\leq a\leq b\leq c$. Then we have to show that $a+b+c\leq \vert a+b-c\vert + \vert a-b+c\vert + \vert -a+b+c\vert$. This is trivial if $a$, $b$, and $c$ are sides of a triangle (possibly degenerate) as the terms inside the absolute values are all non-negative. So it only remains to prove the case that $c>a+b$. In that case we have to show that $a+b+c \leq c-a-b +a+c-b+b+c-a=3c-a-b$ $\iff a+b \leq c$ which holds.

Regarding the generalization to an $n$-dimensional space, one straightforward result is that with $\ell_1$-norm replacing the absolute values in $\mathbb{R}^n$ you get the same inequality. It can be easily proved by adding up the inequalities corresponding to each coordinate.

Answer 4

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.)

Answer 5

We need to prove that $$|x+y|+|x+z|+|y+z|\leq |x|+|y|+|z|+|x+y+z|$$ or

$$\left(|x+y|+|x+z|+|y+z|\right)^2\leq \left(|x|+|y|+|z|+|x+y+z|\right)^2$$ and since $$\sum_{cyc}|x+y|^2=\sum_{cyc}|x|^2+|x+y+z|^2,$$ it's enough to prove that $$\sum_{cyc}|(x+y)(x+z)|\leq\sum_{cyc}\left(|xy|+|x(x+y+z)|\right)$$ or $$\sum_{cyc}|yz+x(x+y+z))|\leq\sum_{cyc}\left(|yz|+|x(x+y+z)|\right),$$ which is just a triangle inequality.

Done!