TITLE: An almost reverse triangle inequality involving two complex numbers
QUESTION [6 upvotes]: Given two complex numbers $u$ and $v$, numerically it seems that
$$1+\vert u\vert +\vert v\vert\le \vert 1+u\vert +\vert 1+v\vert+\vert u+v\vert+\vert 1+u+v\vert\tag{$*$}$$
with equality holding for example when $(u=v=-1)$ Or $(u=\exp(\frac{2i\pi}{3}),v=\exp(\frac{4i\pi}{3}))$    but these are not the only cases.
This inequality does not follow in a clear way to me from Hlawka's inequality, but it looks some how similar to a difficult inequality 
$$\vert u\vert +\vert v\vert\le \vert 1+u\vert +\vert 1+v\vert+\vert1+uv\vert$$
which was proved in this paper.
I would appreciate any idea or help in proving $(*)$.

REPLY [4 votes]: This follows from Hlawka. Denormalizing the inequality, you want to show
$$|x| + |y| + |z| \leq |x+y| + |y+z| + |z+x| + |x+y+z|$$
Substitute $u = y+z, v = z+x, w = x+y$, you want to show
$$|v+w-u| + |w+u-v| + |u+v-w| \leq 2(|u| + |v| + |w|) + |u+v+w|$$
But Hlawka tells you that 
$$\begin{align*}
\text{Right hand side} &\ge (|u| + |v| + |w|) + (|u+v| + |v+w| + |w+u|) \\
&= (|v+w| + |u|) + (|w+u| + |v|) + (|u+v| + |w|) \\
&\ge |v+w-u| + |w+u-v| + |u+v-w| \\
&= \text{Left hand side}
\end{align*}$$