Again Minimum

Applying the Viete's theorem, we have $\left\{\begin{array}{l}a+b+c=1\\ab+bc+ca=\omega^4+4\omega^2+1\\abc=\omega^2\end{array}\right.$

We have:

$\displaystyle \sum_{cyc} \dfrac{a^2+b^2}{ab}=\dfrac{a^2c+b^2c+b^2a+c^2a+c^2b+a^2b}{abc}$

$\qquad\qquad\quad\;=\dfrac{(a+b+c)(ab+bc+ca)-3abc}{abc}$

$\qquad\qquad\quad\;=\dfrac{\omega^4+4\omega^2+1-3\omega^2}{\omega^2}$

$\qquad\qquad\quad\;=\omega^2+\dfrac{1}{\omega^2}+1$

$\qquad\qquad\quad\;\ge2\sqrt{\omega^2.\dfrac{1}{\omega^2}}+1=3$

So, the minimum value is $\boxed{3}$ if and only if $\omega^2=1$ .