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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inequality
complex-numbers
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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
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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.