Call them a,b,c instead for simplicity. A few things of note:
![](https://latex.codecogs.com/png.latex?\bg_white z = x+yi,\ x,y \in \mathbb{R} \implies z\overline{z} = (x+yi)(x-yi) = x^2 + y^2 =|z|^2 )
Also,
![](https://latex.codecogs.com/png.latex?\bg_white \text{If } z \text{ lies on the unit circle, then}\ |z| = z\overline{z} = 1 \implies \overline{z} = \frac{1}{z} )
So,
![](https://latex.codecogs.com/png.latex?\bg_white \begin{align*}\operatorname{LHS}^2= |a+b+c|^2 &= (a+b+c)\overline{(a+b+c)} = (a+b+c)(\overline a+\overline b+ \overline c) = (a+b+c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \\ &= 3 + \frac{a}{b} + \frac{b}{a} + \frac{b}{c} + \frac{c}{b} + \frac{a}{c} + \frac{c}{a} \\ \operatorname{RHS}^2 = |ab+bc+ca|^2 &= (ab+bc+ca)\left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}\right) = 3 + \frac{ab}{bc} + \frac{bc}{ab} + \frac{bc}{ca} + \frac{ca}{bc} + \frac{ab}{ca} + \frac{ca}{ab} \\ &= 3 + \frac{a}{b} + \frac{b}{a} + \frac{b}{c} + \frac{c}{b} + \frac{a}{c} + \frac{c}{a} = \operatorname{LHS}^2 \\ \therefore \operatorname{LHS} &= \operatorname{RHS}\ [\operatorname{LHS}, \operatorname{RHS} \geq 0]\end{align*})
To be honest, we don't need to use the fact that the conjugate is the reciprocal here, but it just made typesetting easier. The key is to use the fact that
then expand and equate