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I asked this question on stackexchange, but despite much effort on my part have been unsuccesful in finding a solution.

Does the inequality $$2(|a|+|b|+|c|) \leq |a+b+c|+|a+b-c|+|a+c-b|+|b+c-a|$$ hold for all complex numbers $a,b,c$ ? For real values a case analysis will verify the inequality. What is desired is a proof using the triangle inequality or a counterexample. Thanks in advance.

Bill Johnson
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Rene Schipperus
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    See answers and comments at http://math.stackexchange.com/questions/793905/inequality-for-absolute-values. – Dietrich Burde May 20 '14 at 12:35
  • Yes that was the question I asked on stackexchange, none of the answers give a proof valid for complex numbers. I posted my own solution but later deleted it (can it still be seen ?) as it contains an error. – Rene Schipperus May 20 '14 at 12:42
  • The first paragraph of the S.B. answer gives a proof by cases for $\mathbb{R}$, next he/she remarks that by adding coordinates one has a proof for $\mathcal{l}_1$ norm, but this is not the same as the Euclidean norm. Am I missing something ? – Rene Schipperus May 20 '14 at 12:52
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    You cannot prove it using just the triangle inequality, because it fails in $\mathbb R^3$ with the $l_\infty$ norm: just take the standard basis vectors for $a,b,c$. You’ll probably need to use that $\mathbb C$ is an inner product space. – Emil Jeřábek May 20 '14 at 13:35
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    Wow that is very insightful. Thank you. – Rene Schipperus May 20 '14 at 13:40
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    Once you have it for $\ell_1^n$ for all $n$ you have it for $L_1(0,1)$ by approximation. Once you have it for $L_1(0,1)$ you have it for Hilbert spaces because $\ell_2$ embeds isometrically into $L_1(0,1)$ (as the span of IID $N(0,1)$ random variables). – Bill Johnson May 20 '14 at 14:22
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    If you want to be more sophisticated, once you have it for some infinite dimensional space you have it for Hilbert spaces by Dvoretzky's theorem. – Bill Johnson May 20 '14 at 14:23
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    Even more sophisticated is that every two dimensional real Banach space embeds isometrically into L$_1(0,1)$, so the inequality is true in all two dimensional Banach spaces. – Bill Johnson May 20 '14 at 14:38
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    Finally, since $L_p$ embeds isometrically into $L_1$ when $1\le p \le 2$ the inequality is true for these Banach spaces. It is false in $\ell_p^3$ for $p>p_0$, where $(3/2)^{p_0}=3$, but what happens when $2< p \le p_0$? – Bill Johnson May 20 '14 at 14:49
  • So to summarize and see if I understand your answer, we first prove the inequality for $\mathbb{R}$, then add $n$ such inequalities together to get the inequality for $\mathbb{R}$ under the $\mathcal{l}_1$ norm, then by approximation we get the same inequality in the function space $L_1(0,1)$ and then imbed $\mathcal{l}_2$ in this space. So we have the relation in any Hilbert space. This is quite an elegant, but sophisticated solution. Thanks. I was looking for something more along the lines of a elementary inequality proof. – Rene Schipperus May 20 '14 at 14:51
  • Right; that is the argument. I did not think about an elementary proof for the plane. – Bill Johnson May 20 '14 at 14:57
  • If we choose the third roots of unity for a,b,c then equality holds. – jjcale May 20 '14 at 19:17
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    The inequality (in the variant given on MSE) is a case $k=3$ of an inclusion-exclusion-like inequality $\sum_{I\subseteq{1,\dots,k}}(-1)^{|I|}\Bigl|\sum_{i\in I}x_i\Bigr|\le0$, which also holds easily for $k=0,1,2$. I don’t really know whether it’s true for larger $k$, but if so, this might suggest where to look for an elementary proof. – Emil Jeřábek May 20 '14 at 19:30
  • Maybe this works?: If $a = b = c$, equality holds. If $a = b \ne c$, then you can assume $a = b = 1$ and the inequality follows by the triangle inequality. Finally, if all three are distinct, then you show that the derivative with respect to any one of them is nonzero. – Deane Yang May 20 '14 at 19:54
  • @Deane Yang : Doesn't work because for the 3 distinct third roots of unity also equality holds. – jjcale May 20 '14 at 20:10
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    For what it's worth, equality is attained whenever $a+b+c=0$. – Seva May 20 '14 at 20:40
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    @EmilJeřábek : I also thought to the inclusion-exclusion generalization, unfortunately it is false for any $k\geq 4$ (and dimension $1$), taking $x_i$ all equal to $1$, except one equal to $-2$. See http://mathoverflow.net/q/184278/6451 and page 174 of Mitrinovic's "Analytic Inequalities" mentioned by Zurab Zilagadze's answer. – BS. Oct 14 '14 at 08:51

3 Answers3

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In general, once you've proven an inequality like this in ${\bf R}$ it holds automatically in any Euclidean space (including ${\bf C}$) by averaging over projections. ("Inequality like this" = inequality where every term is the length of some linear combination of variable vectors in the space; here the vectors are $a,b,c$.) In the case of complex numbers we have $$ |z| = \frac14 \int_0^{2\pi} \bigl| {\rm Re}(e^{i\theta} z) \bigr| \, d\theta. $$ Applying this to $z=a$, $b$, $c$, and $a \pm b \pm c$ reduces the desired inequality to the one-dimensional case. In $d$-dimensional space we'd write $C\|z\|$ as an average of $|u \cdot z|$ over $u$ in the unit sphere (for a suitable constant $C>0$).

I learned this trick at MOP 30+ years ago, and don't know or remember who discovered it. I didn't even know that the specific inequality we were assigned was due to Hlawka (if I remember right that it was the inequality $$ \|x+y\| + \|y+z\|+\|z+x\| \le \|x\|+\|y\| + \|z\| + \|x+y+z\| $$ quoted by Suvrit). We were shown the averaging solution after laboring to prove it bare-handed. The reference Suvrit cites does not use the averaging method, so I do not know whether it too is due to Hlawka or to another mathematician.

Noam D. Elkies
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    Notice that $T:\ell_2^n \to L_1(S^{n-1})$ defined by $(Tx)(y):= \langle x, y \rangle$ is another (multiple of an) isometric embedding of an $n$ dimensional Hilbert space into $L_1$. So at the appropriate conceptual level, the two proofs are basically the same. – Bill Johnson May 21 '14 at 06:35
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    Very nice trick! +1 – Malik Younsi May 21 '14 at 14:09
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It seems that your inequality is just an incarnation of Hlawka's inequality which says that for any vectors $x, y, z$ in an inner product space $V$ we have

\begin{equation*} \|x+y\| + \|y+z\|+\|z+x\| \le \|x\|+\|y\| + \|z\| + \|x+y+z\|. \end{equation*}

Using $x=a+b-c$, $y=a+c-b$, and $z=b+c-a$ we obtain the inequality in the OP.

Additional remarks:

To add some more context, please see the paper linked here, which provides quite a nice summary of work related to Hlawka's inequality, which apparently stems back to a 1942 paper of Hornich (also cited by Zurab below). The paper linked to above explores the interesting generalization: \begin{equation*} f(x+y) + f(y+z) + f(z+x) \le f(x+y+z) + f(x)+f(y)+f(z), \end{equation*} where $x,y,z$ may come from an Abelian group, or a linear space, or the real line---each with its own set of conditions on the mapping $f$. The functional form of Hlawka's inequality is credited to a 1978 paper of Witsenhausen.

Suvrit
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  • Do you know a proof of it? – Qiaochu Yuan May 21 '14 at 00:46
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    For a proof please see: http://books.google.com/books?id=UWB4I8mugOQC&lpg=PA100&dq=hlawka%20inequality&pg=PA100#v=onepage&q=hlawka%20inequality&f=false . That link also mentions extension to Banach spaces by Lindenstrauss and Pelcynski (under certain embedability assumptions) – Suvrit May 21 '14 at 00:50
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    It looks like Lindenstrauss & Pelczynski had in mind the same observations I made in comments above. That approach for extending inequalities from the real line to $L_p$ spaces has of course been around for a long time. – Bill Johnson May 21 '14 at 04:28
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In fact the Hlawka's inequality first appeared (as a special case of more general result) in H. Hornich, Eine Ungleichung für Vektorlängen, Mathematische Zeitschrift 48 (1942), 268-274 http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=PPN266833020_0048&DMDID=DMDLOG_0025&LOGID=LOG_0025&PHYSID=PHYS_0256 (see p. 268. P.S. as Joni Teräväinen has remarked, Hornich credits on page 274 to Hlawka an algebraic proof of this special case and reproduces it).

Hlawka's original proof, besides the book indicated by Suvrit, can be found in "Classical and New Inequalities in Analysis" by D.S. Mitrinovic, J. Pecaric and A.M Fink, p. 521 and in "Analytic Inequalities" by D.S. Mitrinovic, p.171. Both books provide Adamovic and Djorkovic generalizations of the Hlawka's inequality. Interestingly, all these generalizations are special cases of more general result given in http://www.sciencedirect.com/science/article/pii/S0022247X96904588 (Generalizations of Dobrushin's Inequalities and Applications, by M. Radulescu and S. Radulescu).

Another proof of Hlawka's inequality can be found in http://www.sbc.org.pl/Content/34160/1995_13.pdf (On two geometric inequalities, by A. Simon, P. Volkmann), and still another one in http://www.jstor.org/discover/10.2307/2310890?uid=3738936&uid=2&uid=4&sid=21104051771107 (The Polygonal Inequalities, by D.M. Smiley and M.F. Smiley).

Zurab Silagadze
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    On page 274 it says that Hlawka had already proved the special case which is this inequality. I'm not sure though if that was published. – Joni Teräväinen May 21 '14 at 11:38