Finding the minimum value of $|a+b\omega+c\omega^2|$ if $a,b,c$ are unequal integers where $\omega^3=1$ My try 1:
$$|a+b\omega+c\omega^2|\le\sqrt{|a+b+c||\underbrace{1+\omega+\omega^2}_0|}$$
Cauchy-Scwartz won't give us an upper bound since $a,b,c$ are nonequal integers.

My try 2:
$$|a+b\omega+c\omega^2|=\left|a+b\left(\frac{-1+i\sqrt3}{2}\right)+c\left(\frac{-1-i\sqrt3}{2}\right)\right|\\
=\sqrt{\left(a-\frac b2-\frac c2\right)^2+\frac34\left(b-c\right)^2}
\\=\sqrt{a^2+b^2+c^2-ab-bc-ca}
\\=\frac1{\sqrt2}\sqrt{(a-b)^2+(b-c)^2+(c-a)^2}$$

So it's just a kind of measure of dispersion of the numbers, a most suitable choice would be then the triplet $\{-1,0,1\}$ which gives an upper bound of $\sqrt3$ but it's not quite close (there's a more nearer one i.e.1)
Geometrically, the set 
$$X =\{\; a + b\omega + c \omega^2 : (a,b,c) \in \mathbb{Z}^3\;\}$$
forms a triangular lattice in $\mathbb{C}$. Consider its subset 
$$Y = \{\; a + b\omega + c \omega^2 : (a,b,c) \in \mathbb{Z}^3, a \ne b \ne c \ne a\;\}$$
For any complex number $z \ne 0$, let $\mathbb{R}z$ be the line $\{\; \lambda z : \lambda \in \mathbb{R} \;\}$ passing through $z$ and the origin.
It is easy to see

$$\begin{array}{ccr}
a \ne b &amp;\implies&amp; Y \cap \mathbb{R}\omega^2 = \emptyset\\
b \ne c &amp;\implies&amp; Y \cap \mathbb{R} = \emptyset\\
c \ne a &amp;\implies&amp; Y \cap \mathbb{R}\omega = \emptyset\\
\end{array}
$$
After we remove $\mathbb{R}, \mathbb{R}\omega, \mathbb{R}\omega^2$ from $X$, the closest candidate that can belong to $Y$ all lies at a distance $\sqrt{3}$ from $0$.
Since one of these candidates $1 - \omega^2$ can be rewritten as $3 + 2 \omega + \omega^2$, it belongs to $Y$ hence the minimum we seek is $\sqrt{3}$.
$\bf{My\; Solution::}$ Let $$\displaystyle f(a,b,c) = \left|a+b\omega+c\omega^2\right|\Rightarrow f^2(a,b,c) = \left|a+b\omega+c\omega^2\right|^2 $$

$$\displaystyle = \left(a+b\omega+c\omega^2\right)\cdot\left(a+b\omega^2+c\omega\right) $$

Above we have used the fact $$\left|z\right|^2=z\cdot \bar{z}$$ and We take $$\left[f(a,b,c)\right]^2 = f^2(a,b,c).$$  

So We Get $$\displaystyle f^2(a,b,c)=a^2+ab\omega+ac\omega^2+ab\omega^2+b^2\omega^3+bc\omega^4+ac\omega+bc\omega^2+c^2\omega^3$$

Now Using the fact $$1+\omega+\omega^2 = 0$$ and $$\omega^3=1$$

We Get $$\displaystyle f^2(a,b,c) = a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$$

Now Given $a,b,c$ are Three Different Integers, and Diff. of Two Integer is Minimum 

when we take as $a=n-1\;,b=n$ and $c=n+1$

We Get $$\displaystyle f^2(a,b,c) = \frac{1}{2}\left[1^2+1^2+2^2\right]=3$$

So We get $$\displaystyle f^2(a,b,c) = 3\Rightarrow f(a,b,c) = \sqrt{3}$$.

If $a,b$ and $c$ are $3$ unequal  Integers.