# Show that $|v| + |v + w| + |w + z| + |2 + z|\ge 2$ for complex numbers $v,w,$ and $z$

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.

$$|v + (-v-w)+(w+z)+(-z-2)| \le |v|+|v+w|+|w+z|+|z+2|$$
• @martycohen Right, and the implied $0$ as well: $|a+v| + |v + w| + |w + z| + |z+b| \ge |a-b|$. – dxiv Jul 9 '18 at 1:48