Show $|a|+|b|+|c|+|a+b+c| \geq  |a+b|+|b+c|+|c+a|$ for complex $a$, $b$, $c$ 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?

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. 
