# 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.

• Also, what absolute value inequalities do you know already? – grantfgates May 14 '14 at 1:10
• The comments to the MO post do not show that the inner product properties must be used, rather only that the triangle inequality alone is insufficient. Of course, it seems entirely reasonable that the inner product will play a roll. – RghtHndSd May 20 '14 at 16:59

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.)

• Suvrit also observed in his answer to the mathoverflow question that this particular inequality is due to Hlawka, in the form $$\|x+y\| + \|y+z\|+\|z+x\| \le \|x\|+\|y\| + \|z\| + \|x+y+z\|$$ (with $(x,y,z) = (a+b-c, a+c-b, b+c-a)\,$). – Noam D. Elkies May 21 '14 at 16:12

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 $(2)$.

• Thanks, I suspected there was a proof along these lines. However there is the further question of whether ot holds for complex numbers, or more generally in n dimensions. I have not been able to find a counterexample. – Rene Schipperus May 14 '14 at 1:09
• I can provide empirical evidence that it holds in the complex numbers but I cannot provide a proof. – Brad May 14 '14 at 1:16
• I guess the real question is can you get a proof using the triangle inequality, rather than a case analysis. – Rene Schipperus May 14 '14 at 1:21
• I have proven it using the triangle inequality. – Brad May 14 '14 at 2:12
• I am hoping that some such approach will work, but I dont see how the argument above will succeed. One can Say $$4a=(a+b+c)+(a+b-c)+(a-b+c)+(a-b-c)$$ with similar equations for $b$ and $c$, then using triangle inequalitiy and adding one gets – Rene Schipperus May 14 '14 at 9:47

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 $LHS = x - 1 + y - 1 + 1 - z = x + y - z - 1 \leq |x + y - z - 1| \leq |x| + |y| + |-z| + |-1| = 1+ |x| + |y| + |z| = RHS$

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.

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!