A proof about complex number If $a, b, c\in \mathbb{C}$, and if $\left \| a \right \|=\left \| b \right \|=\left \| c \right \|=1$, prove $(a+b)(b+c)(c+a)/(abc)\in \mathbb{R}$.
I have thought this Q for a long time, but I can only get something long and troublesome but not the answer. Can anyone help me please? THANK YOU! ~
 A: Since $|a|=|b|=1$, we have that
$$
\left|\frac{a}{b}\right|=1 \quad \Rightarrow \quad \overline{\dfrac{a}{b}} =\left.\left|\dfrac{a}{b}\right|^2 \middle/\dfrac{a}{b}\right. =\frac{b}{a}
$$
Therefore,
$$
\begin{align}
\frac{(a+b)^2}{ab}
&=\frac{a}{b}+2+\frac{b}{a}\\
&=2+2\,\mathrm{Re}\left(\frac{a}{b}\right)\\
&\ge0
\end{align}
$$
This means that
$$
\left(\frac{(a+b)(b+c)(c+a)}{abc}\right)^2=\frac{(a+b)^2}{ab}\frac{(b+c)^2}{bc}\frac{(c+a)^2}{ca}\ge0
$$
which leads immediately to
$$
\frac{(a+b)(b+c)(c+a)}{abc}\in\mathbb{R}
$$
A: Here's a nice geometric argument (with some holes and assumptions left to fill up):
\begin{align}
\arg \frac {(a+b)(b+c)(a+c)}{abc} &= \arg (a+b)+\arg(b+c)+\arg(a+c) - \arg (a) -\arg( b) - \arg( c) \\ &= \frac{\arg (a)+arg(b)} 2 + \frac{\arg (b)+arg(c)} 2 + \frac{\arg (a)+arg(c)} 2 \\ & \qquad\qquad - \arg (a) -\arg( b) - \arg( c) \\&= 0
\end{align}
A: Let us denote the quantity by $q$.  Since $a$, $b$ and $c$ are all of unit modulus, 
$${\overline q} = \overline{(a+b)(a+c)(b+c)/(abc)} = (1/a + 1/b)(1/a + 1/c)(1/b + 1/c)(abc) $$
now add the fractions and get
$${\overline q} = (abc){a + b\over ab}{a + c\over ac}{b + c\over bc} = (a+b)(b+c)(a+c)/(abc) = q.$$
A: Hint 1.
Since $\Vert a\Vert=\Vert b\Vert=\Vert c\Vert=1$, then 
$$
a=e^{i\alpha}\qquad
b=e^{i\beta}\qquad
c=e^{i\gamma}\tag{1}
$$
Hint 2
A complex number $z\in\mathbb{C}$ is real iff 
$$
z+\overline{z}=2z\tag{2}
$$
Substitute $(1)$ into $(2)$ and check that it is correct.
