Let $v, w,$ and $z$ be three complex numbers. Show that $$|v| + |v + w| + |w + z| + |2 + z|\ge 2$$

I thought about squaring both sides but things would get too messy. I tried using the fact that the absolute value of a complex number squared equals it's product with it's conjugate, but I wasn't sure how to proceed.


Hint:   by the triangle inequality:

$$ |v + (-v-w)+(w+z)+(-z-2)| \le |v|+|v+w|+|w+z|+|z+2| $$

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  • $\begingroup$ And of course you can replace 2 by any other value. $\endgroup$ – marty cohen Jul 9 '18 at 1:35
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    $\begingroup$ @martycohen Right, and the implied $0$ as well: $|a+v| + |v + w| + |w + z| + |z+b| \ge |a-b|$. $\endgroup$ – dxiv Jul 9 '18 at 1:48

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