Three complex numbers of modulus one

1. Since $\bar a=1/a$, etc.,
$$\overline{a+b+c}=\frac1a+\frac1b+\frac1c=\frac{ab+bc+ca}{abc}.$$
Taking moduli and using $|abc|=1$: $|a+b+c|=|ab+bc+ca|$.

2. If $a+b+c=0$, part 1 gives $ab+bc+ca=0$. Then $a,b,c$ are the roots of $t^3-abc=0$. The equation $t^3=abc$ has three distinct roots (since $abc\ne0$): the three cube roots of $abc$, which are $r,r\omega,r\omega^2$ — vertices of an equilateral triangle centred at $0$.

3. Let $Z=\dfrac{(a+b)(b+c)(c+a)}{abc}$. For $a,b\ne0$: $\overline{a+b}=\frac1a+\frac1b=\frac{a+b}{ab}$. Therefore
$$\overline{(a+b)(b+c)(c+a)}=\frac{(a+b)(b+c)(c+a)}{(abc)^2},\qquad\overline{abc}=\frac1{abc}.$$
Hence $\bar Z=\dfrac{(a+b)(b+c)(c+a)}{(abc)^2}\cdot abc=Z$, so $Z$ is real. $\blacksquare$

Lesson. On the unit circle replace $\bar z$ by $1/z$: symmetric expressions become rational functions you can compare with their conjugates.