If a, b and c are distinct integers and $\omega$($\ne$1) is a cube root of unity, then the minimum value of $\left|a+b\omega +c{\omega }^2\right|+\left|a+b{\omega }^2+c\omega \right|$, is

Let $z=a+b\omega +c{\omega }^2$. Then,

$\begin{array}{l}|z{|}^2=z\overline{z}=\left(a+b\omega +c{\omega }^2\right)\left(a+b\overline{\omega }+c\overline{{\omega }^2}\right)\\ =\left(a+b\omega +c{\omega }^2\right)\left(a+b{\omega }^2+\mathrm{c}\omega \right)\\ =a^2+b^2+c^2-ab-bc-ca\\ =\frac{1}{2}\left[(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\right]\\ \Rightarrow |z{|}^2\ge \frac{1}{2}\times 6=3 \left[\begin{array}{l}\because a\ne b\ne c\\ \therefore |a-b|\ge 1,|b-c|\ge 1\\ \text{ and }|a-c|\ge 2\end{array}\right] \end{array}$

$\begin{array}{l}\therefore \left|z\right|\ge \sqrt{3}\\ \begin{array}{ll}\therefore \left|a+b\omega +c{\omega }^2\right|& +\left|a+b{\omega }^2+c\omega \right|\end{array}\\ \begin{array}{l}=\left|a+b\omega +c{\omega }^2\right|+\left|a+b\overline{{\omega }^2}+c\overline{\omega }\right|\\ =\left|a+b\omega +\alpha {\omega }^2\right|+\left|a+b\omega +\mathrm{c}{\omega }^2\right|\\ =2\left|a+b\omega +c{\omega }^2\right|=2\left|z\right|\ge 2\sqrt{3}\end{array}\end{array}$

Hence, the minimum value of $\left|a+b\omega +\mathrm{c}{\omega }^2\right|+\left|a+b{\omega }^2+\mathrm{c}\omega \right|$ is $2\sqrt{3}$.