Let $A,B,C$ be points on complex plane corresponding to $\alpha_1, \alpha_2, \alpha_3$, respectively. The condition implies that $A,B,C$ forms a non-degenerate triangle, and its centroid $G$ is the origin. The original inequality is thus interpreted as a geometric inequality: \[ \sum_{cyc} \frac{|BC|}{\sqrt{|GA|}} \left( \frac{1}{\sqrt{|GB|}}+\frac{1}{\sqrt{|GC|}}-\frac{2}{\sqrt{|GA|}} \right) \le 0. \]Simple AM-GM gives $\frac{1}{\sqrt{|GA|}\sqrt{|GB|}} \le \frac{1}{2} \left( \frac{1}{|GA|}+\frac{1}{|GB|} \right)$ and so on, therefore we only need to prove \[ \sum_{cyc} |BC| \left( \frac{1}{|GB|}+\frac{1}{|GC|}-\frac{2}{|GA|} \right) \le 0, \]or \[ \left( |BC|+|CA|+|AB| \right) \left( \frac{1}{|GA|}+\frac{1}{|GB|}+\frac{1}{|GC|} \right) \le 3 \left( \frac{|BC|}{|GA|}+\frac{|CA|}{|GB|}+\frac{|AB|}{|GC|} \right). \]Without loss of generality, let $|AB| \ge |AC| \ge |BC|$, then it follows that $|GA| \ge |GB| \ge |GC|$, so the sequences in two brackets are in same order, hence by Chebyshev's inequality (or rearrangement inequality) we are done. As for the equality case, AM-GM already requires $|GA|=|GB|=|GC|$, so $|AB|=|BC|=|CA|$, i.e. $\alpha_1, \alpha_2, \alpha_3$ forms an equilateral triangle centered at origin.