<s>[INST] Prove equilateral triangle if vertices $a,b,c$ satisfy $|a|=|b|=|c|$ and $\sum_{cyc}\frac1{8-a/b-b/a-a/c-c/a}=0.3$ 
If $a,b,c\in \mathbb{C}$ are different complex numbers in pairs and $|a|=|b|=|c|=1$ and $P(a),Q(b),R(c)$ are the vertices of a triangle and $\displaystyle \sum_{\bf{cyc}}\frac{1}{8-\frac{a}{b}-\frac{b}{a}-\frac{a}{c}-\frac{c}{a}}=0.3.$ Then show that $PQ=QR=RP.$

What I tried
Let $a=e^{i\alpha}=\cos \alpha+i\sin \alpha$
and $b=e^{i\beta}=\cos \beta+i\sin \beta$
and $c=e^{i\gamma}=\cos \gamma+i\sin \gamma$
Now $\displaystyle \frac{a}{b}+\frac{b}{a}=e^{i(\alpha-\beta)}+e^{-i(\alpha-\beta)}=2\cos (\alpha-\beta)$
and $\displaystyle \frac{a}{c}+\frac{c}{a}=e^{-i(\gamma-\alpha)}+e^{i(\gamma-\alpha)}=2\cos(\gamma-\alpha)$
$\displaystyle \Longrightarrow \sum_{\bf{cyc}}\frac{1}{8-2\cos(\alpha-\beta)-2\cos(\gamma-\alpha)}=0.3$
$\displaystyle \Longrightarrow \sum_{\bf{cyc}}\frac{1}{4-\cos(\alpha-\beta)-\cos(\gamma-\alpha)}=0.6$
Let $\alpha-\beta=A,\beta-\gamma=B,\gamma-\alpha=C$
$\displaystyle \Longrightarrow \sum_{\bf{cyc}}\frac{1}{4-\cos A-\cos C}=0.6$
How do  I solve it after that
Please help. 
 [/INST] Apply the Jensen's inequality 
$$f(x_1) + f(x_2) + f(x_3) \ge 3f\left(\frac{x_1+x_2+x_3}3\right)$$
for the convex function $f(x)=\frac1x$ to get
$$0.3=\sum_{{cyc}}\frac{1}{8-\frac{a}{b}-\frac{b}{a}-\frac{a}{c}-\frac{c}{a}}
\ge \frac{1}{24-2\left(\frac{a}{b}+\frac{b}{a}+\frac bc+\frac cb+\frac{c}{a}+\frac{a}{c}\right)}
$$
or,
$$\frac{a}{b}+\frac{b}{a}+\frac bc+\frac cb+\frac{c}{a}+\frac{a}{c}\le -3\tag 1$$
Given that $|a|=|b|=|c|$, assume, without loss of generality $\alpha, \ \beta, \ \gamma >0$,
$$\frac ba =e^{i\alpha},\>\>\>\>\>\frac cb =e^{i\beta},\>\>\>\>\>\frac ac=e^{i\gamma}
\>\>\>\>\>\alpha + \beta + \gamma = 2\pi$$
Then, the inequality (1) becomes,
$$\cos\alpha + \cos\beta + \cos \gamma \le -\frac32\tag 2$$
Note that,
$$\cos\alpha + \cos\beta + \cos \gamma
= 2\cos\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2 + 2\cos^2\frac\gamma2-1$$
$$= 2\cos^2\frac\gamma2 - 2\cos\frac{\gamma}2\cos\frac{\alpha-\beta}2 -1
\ge 2\cos^2\frac\gamma2 - 2\cos\frac{\gamma}2 -1 $$
$$= 2\left(\cos\frac\gamma2 - \frac12\right)^2 -\frac32 \ge-\frac32\tag 3$$
From (2) and (3), we have 
$$\cos\alpha + \cos\beta + \cos \gamma=-\frac32$$
and, according to (3), the equality occurs at $\cos\frac\gamma2 - \frac12=0$ and $\cos\frac{\alpha-\beta}2 =1$, which leads to $\alpha=\beta =\gamma = 120^\circ$. Thus, the neighboring vertexes of equal module all have the same argument angle between them, hence, forming an equilateral triangle.
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