Prove equilateral triangle if vertices $a,b,c$ satisfy $|a|=|b|=|c|$ and $\sum_{cyc}\frac1{8-a/b-b/a-a/c-c/a}=0.3$

Question

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.

Answer 1

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.