How to prove for any complex numbers $a$, $b$, $c$, the inequality $$|a|+|b|+|c|+|a+b+c| \geq |a+b|+|b+c|+|c+a|$$ is correct?
Show $|a|+|b|+|c|+|a+b+c| \geq |a+b|+|b+c|+|c+a|$ for complex $a$, $b$, $c$
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1I changed the variables to $a$, $b$, and $c$ from $z_1$, $z_2$, and $z_3$ so that the title would fit on one line on the front page, but if that's not acceptable to you you can change it back. – 2012-06-24
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1I feel there should be a $n$-variable version, like inclusion-exclusion formula: $$|a|+|b|+|c|+|d|-|a+b|-|a+c|-\dots-|c+d|+|a+b+c|+|b+c+d|+|c+d+a|+|d+a+b|-|a+b+c+d| \geq 0$$ but I'm unable to prove it. – 2012-06-24
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0Good generalization,I think MGNewman's method would be useful to proof that. – 2012-06-25
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1what a pity,the situation for n = 4 is wrong,let a=2,b=-1,c=-1,d=-1. – 2012-06-25
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0@sdcvvc There is a $n$-dimensional version: Prove exactly the same inequality, but when $a$, $b$ and $c$ are vectors in $\mathbf R^n$. – 2018-04-12
2 Answers
Both sides are non-negative, so it suffices to show that the square of the left-hand-side is at least the square of the right-hand-side. That is, we wish to show: $$ |a|^2+|b|^2+|c|^2+|a+b+c|^2+2|ab|+2|bc|+2|ac+2(|a|+|b|+|c|)|a+b+c| \geq\\ |a+b|^2+|b+c|^2+|a+c|^2+2(|a(a+b+c)+bc|+|b(a+b+c)+ac|+|c(a+b+c)+bc|) $$ The square terms cancel: $$ |a|^2+|b|^2+|c|^2+|a+b+c|^2 = 2|a|^2+2|b|^2+2|c|^2+2\operatorname{Re}(ab+bc+ac)=|a+b|^2+|b+c|^2+|a+c|^2 $$ and by the triangle inequality we have $|a(a+b+c)|+|bc|\geq |a(a+b+c)+bc|$ and cyclic permutations.
Here is a more geometric point of view. But it is clearly less efficient then the other answer for this special case. However the proof can be easily extended for any number of points.
$a,b$ and $c$ are seen as vertices of a triangle.
Let $a=z+a_0,b=z+b_0$ and $c=z+c_0$ with $a_0+b_0+c_0=0$, $z$ is the orthocenter of the triangle.
The LHS of the inequality we want to prove rewrites as :
$$|a|+|b|+|c|+|a+b+c|=|z+a_0|+|z+b_0|+|z+c_0|+|3z|.$$
On the other hand :
$$|a+b|+|b+c|+|c+a|=|2z+a_0+b_0|+|2z+b_0+c_0|+|2z+c_0+a_0|.$$
We then derive :
$$|a+b|+|b+c|+|c+a|\le|3z|+|z-c_0|+|z-a_0|+|z-b_0|.$$
We are then left to prove that :
$$|z+a_0|+|z+b_0|+|z+c_0| \ge |z-c_0|+|z-a_0|+|z-b_0|.$$
Which is the same as :
\begin{align} |a|+|b|+|c| & \ge |c-2c_0|+|a-2a_0|+|b-2b_0|\\ & \ge |c|-2|c_0|+|a|-2|a_0|+|b|-2|b_0| \\ & \ge |c|+|a|+|b|-2|c_0+a_0+b_0|\\ & = |a|+|b|+|c|. \end{align}
The second inequality is due to the second triangular inequality.
Then the result is proved. Note that we can easily generalize the inequality and that proof for $n$ complex points. The points being seen as the vertices of a (may be self-intersecting) $n$-gon and $z$ the barycenter of those points. The corresponding generalized inequalities would be :
$$\sum_{i=1}^n |a_i| + \left|\sum_{i=1}^n a_i \right| \ge \sum_{i=1}^n |a_i+a_{i+1}+\dots+a_{i+n-1}|,$$
Where $a_{i+n}:=a_i$ for every $i \in \{1,\dots,n\}$.
Note that any permutation of the $a_i$ yields the same result.