Let Z , W , U , V complex numbers. Prove $|Z|+|U|+|V|+|W| ≤ |Z+U|+|Z+V|+|Z+W|+|U+V|+|U+W|+|V+W|$

Question

Let z , w , u , v complex numbers. Prove

$|z|+|u|+|v|+|w| ≤ |z+u|+|z+v|+|z+w|+|u+v|+|u+w|+|v+w|$

My try: $2z = (z+w) − (w+v) + (v+z)$

$2w= (w+u) − (u+z) + (z+w)$

$2u = (u+w) − (w+v) + (v+u)$

$2v = (v+u) − (u+z) + (z+v)$

But I don't knew what next.

Answer 1

By triangle inequality, we have \begin{align} 2|z|&=|2z|\leq |z+w|+|w+v|+|v+z| \\ 2|w|&=|2w|\leq |w+u|+|u+z|+|z+w| \\ &... \end{align} and add up these 4 inequalities.