Problem complex number inequality:
Given $x,y,z$ be complex numbers such that $|x|, |y|, |z| \le 1$ and $|x+y+z| \le 1.$ Prove that $$|(x+y)(y+z)(z+x)| \le 1.$$
I see it on AOPS here.
Here is my attempt:
Let $(m;n;p)=\left(\dfrac{x}{x+y+z}; \dfrac{y}{x+y+z}; \dfrac{z}{x+y+z}\right)$ then $|m|, |n|, |p| \le 1$ and $m+n+p=1.$ On the other hand:
$$|(x+y)(y+z)(z+x)| \le 1 \Leftrightarrow |(m+n)(n+p)(p+m)| \le 1.$$
$\bullet$ Case 1: At least one in three numbers $m,n,p$ is real. Assume it is $m.$
Hence, $\exists a,b,c \in \mathbb{R} : n=a+ci; p=b-ci$ and $m=1-n-p=1-a-b.$
Since $|m|, |n|, |p| \le 1$ then $a^2+c^2 \le 1; b^2+c^2 \le 1$ and $0 \le a+b \le 2.$
We have: \begin{align*} |(m+n)(n+p)(p+m)| &=|1-b+ci|\cdot|1-a-ci|\cdot|a+b|\\ &=\sqrt{1-2b+b^2+c^2}\cdot\sqrt{1-2a+a^2+c^2}\cdot (a+b)\\ &\le \sqrt{2(1-b)\cdot 2(1-a)}\cdot (a+b)\\ &=2\sqrt{(1-a)(1-b)}\cdot (a+b)\\ &\le (2-a-b)(a+b) \le 1.\end{align*}
$\bullet$ Case 2: All $m,n,p$ are not real numbers. I haven't proved it yet.
Comment: In AOPS. There is a solution use strong skill. It let two in three numbers $m,n,p$ have modun $=1$, on the other hand: $|m|=|n|=1$ anh $m+n+p=1.$ If we have this then problem becomes very easy. However, there isn't proof for that, and I haven't proved it yet, also there is no reference about it.