, 2021

Consider the triangle $A_1A_2A_3$ having these three points as vertices, where $A_1(\alpha_1), A_2(\alpha_2), A_3(\alpha_3)$. By the given condition we know that its centroid $O$ is at the origin; let $s_i = |\alpha_{i+1} - \alpha_{i+2}|$ be the side length of $\overline{A_{i+1}A_{i+2}}$. Then, $m_i = \frac{3}{2}|\alpha_i|$ is the length of the median to $\overline{A_{i+1}A_{i+2}}$. Thus (*) is equivalent to $\sum_{i=1}^3 \frac{s_i}{\sqrt{m_i}} \left( \frac{1}{\sqrt{m_{i+1}}} + \frac{1}{\sqrt{m_{i+2}}} - \frac{2}{\sqrt{m_i}} \right) \le 0$, which after rearrangement gives the equivalent inequality:
$$\sum_{i=1}^{3} \frac{s_i + s_{i+1}}{\sqrt{m_i m_{i+1}}} \le 2 \sum_{i=1}^{3} \frac{s_i}{m_i}, \quad (**)$$
where $(s_4, m_4, s_5, m_5) = (s_1, m_1, s_2, m_2)$. We will prove that the following inequality holds:
$$\frac{s_i + s_{i+1}}{\sqrt{m_i m_{i+1}}} \le \frac{s_i}{m_i} + \frac{s_{i+1}}{m_{i+1}}, \quad i = 1, 2, 3. \quad (\#)$$
Proof of (\#): First, squaring both sides of inequality (\#) and rearranging, we obtain an equivalent inequality
$$m_i m_{i+1} (s_i^2 + s_{i+1}^2) \le m_{i+1}^2 s_i^2 + m_i^2 s_{i+1}^2, \quad i = 1, 2, 3. \quad (\#_1)$$
We will show that equality in (\#1) holds only when $s_i = s_{i+1}$.
Step 1. Note that $m_i = \frac{\sqrt{2(s_{i+1}^2 + s_{i+2}^2) - s_i^2}}{2}$. We first give an upper bound estimate for $m_i m_{i+1}$:
$$\begin{aligned} (4m_i m_{i+1})^2 &= (2s_{i+1}^2 + 2s_{i+2}^2 - s_i^2)(2s_{i+2}^2 + 2s_i^2 - s_{i+1}^2) \\ &= (4s_{i+2}^4 + s_{i+1}^2 s_i^2) + 2(s_i^2 + s_{i+1}^2)s_{i+2}^2 - 2(s_{i+1}^2 - s_i^2)^2 \\ &= (2s_{i+2}^2 + s_{i+1}s_i)^2 + 2(s_i - s_{i+1})^2 s_{i+2}^2 - 2(s_{i+1}^2 - s_i^2)^2 \\ &= (2s_{i+2}^2 + s_{i+1}s_i)^2 + 2(s_i - s_{i+1})^2 (s_{i+2} + s_{i+1} + s_i)(s_{i+2} - s_{i+1} - s_i) \\ &\le (2s_{i+2}^2 + s_{i+1}s_i)^2. \end{aligned}$$
Therefore,
$$4m_i m_{i+1} \le 2s_{i+2}^2 + s_{i+1} s_i \quad (\#2)$$
(equality holds if and only if $s_i = s_{i+1}$.)
Step 2. From (\#2) we obtain
$$\begin{aligned} & 4m_{i+1}^2 s_i^2 + 4m_i^2 s_{i+1}^2 - 4m_i m_{i+1} (s_i^2 + s_{i+1}^2) \\ & \ge (2s_{i+2}^2 + 2s_i^2 - s_{i+1}^2)s_i^2 + (2s_{i+1}^2 + 2s_{i+2}^2 - s_i^2)s_{i+1}^2 - (2s_{i+2}^2 + s_{i+1}s_i)(s_i^2 + s_{i+1}^2) \\ & = (2s_i^2 - s_{i+1}^2)s_i^2 + (2s_{i+1}^2 - s_i^2)s_{i+1}^2 - s_{i+1}s_i(s_i^2 + s_{i+1}^2) \\ & = 2(s_i^2 - s_{i+1}^2)^2 - s_{i+1}s_i(s_i - s_{i+1})^2 \\ & = (s_i - s_{i+1})^2 [2(s_i + s_{i+1})^2 - s_{i+1}s_i] \ge 0. \end{aligned}$$
This establishes (\#1). Hence we also obtain (#). Since (**) is obtained by the cyclic sum of (#), and () is equivalent to (**), we thus obtain (*), where equality holds only when $\alpha_1, \alpha_2, \alpha_3$ are the vertices of an equilateral triangle with centroid at the origin $O$ (for example: $\alpha_{i+1}/\alpha_i = e^{2i\pi\sqrt{-1}/3}, i = 1, 2$.)