Bump! I have asked lots of my friends none of them solved it. So bump!

One uses computer maybe the answer is $3888^{1/6}$?

Official ans:$3888^{1/6}$

Here's a failed attempt in my try to revive this topic. We have: $\alpha + \beta + \gamma = 0$ $\alpha\beta + \beta\gamma + \gamma\alpha = a$ $\alpha\beta\gamma=1$ and $\frac{\alpha}{\beta} + \frac{\beta}{\gamma} + \frac{\gamma}{\alpha}=-b$ $\frac{\alpha}{\gamma} + \frac{\beta}{\alpha} + \frac{\gamma}{\beta}=c$ Combining fractions for each of the latter 2 sums and then adding both together gives $|-b+c|=|-3\alpha\beta\gamma|=3$ by using the identity $(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$. I have no progress after here. Some points of contention/remarks from me here: 1. Even if $b$ is negative, I obtain the value $|b+c|=3$. But considering $\frac{|b|+|c|}{a} \geq \frac{|b+c|}{a} = \frac{3}{a} \geq 1$ by AM-GM (on the roots of the second equation summed two-by-two) doesn't give the desired bound. I suspect we can do something more with the value of $a$, but I don't know what to do with it. 2. I suspect we may be able to write the numerator of the desired form in terms of $a$, although again, I don't know how. 3. The modulus signs may motivate us to consider squaring each of $b$ and $c$ separately, although I didn't find significant progress with it. 4. The sixth power in the official answer could also hint at squaring both cubics, although that is very scary. Hopefully this suffices to jumpstart another discussion of this interesting problem

Bumping!

Bump bump

I have reduced it to the problem of minimising $\frac{4(-\alpha^9-3\alpha^6+6\alpha^3-1)}{\alpha(\alpha^3+1)^2}$ for $\alpha\in(-1,0)$. Since $a$ is a positive real number, thus $x^3+ax+1$ is increasing, and this means that there is exactly one real root. WLOG it is $\alpha$, then $\beta$ and $\gamma$ are complex conjugates. Now $$\overline{b}=-\overline{\left(\frac{\alpha}{\beta}+\frac{\beta}{\gamma}+\frac{\gamma}{\alpha}\right)}=-\left(\frac{\alpha}{\gamma}+\frac{\gamma}{\beta}+\frac{\beta}{\alpha}\right)=-c$$This also implies that $|b|=|c|$. We just need to minimise $\frac{4|b|^2}{a}$. We know that $$|b|^2=b\overline{b}=\left(\frac{\alpha}{\beta}+\frac{\beta}{\gamma}+\frac{\gamma}{\alpha}\right)\left(\frac{\alpha}{\gamma}+\frac{\gamma}{\beta}+\frac{\beta}{\alpha}\right)=3+\sum_{cyc}\frac{\alpha\beta}{\gamma}+\sum_{cyc}\frac{\gamma}{\alpha\beta}=3+\sum_{cyc}\alpha^3\beta^3-\sum_{cyc}\alpha^3$$ We also have $\alpha^3+a\alpha+1=0$, and $\alpha+\beta+\gamma=0$. Thus, we can compute $\sum_{cyc}\alpha^3=-3$. We also get that $a=-\frac{\alpha^3+1}{\alpha}$, and since $a$ is a positive real number, this means $\alpha\in(-1,0)$. Also, $$\sum_{cyc}\alpha^3\beta^3=\alpha^3(\beta^3+\gamma^3)+\beta^3\gamma^3=\alpha^3(-3-\alpha^3)-\frac{1}{\alpha^3}$$ Thus, $$\left(\frac{|b|+|c|}{a}\right)^2=\frac{4|b|^2}{a^2}=\frac{4\left(6+\alpha^3(-3-\alpha^3)-\frac{1}{\alpha^3}\right)}{\left(\frac{\alpha^3+1}{\alpha}\right)^2}=\frac{4(-\alpha^9-3\alpha^6+6\alpha^3-1)}{\alpha(\alpha^3+1)^2}$$

ACGNmath wrote: I have reduced it to the problem of minimising $\frac{4(-\alpha^9-3\alpha^6+6\alpha^3-1)}{\alpha(\alpha^3+1)^2}$ for $\alpha\in(-1,0)$. Since $a$ is a positive real number, thus $x^3+ax+1$ is increasing, and this means that there is exactly one real root. WLOG it is $\alpha$, then $\beta$ and $\gamma$ are complex conjugates. Now $$\overline{b}=-\overline{\left(\frac{\alpha}{\beta}+\frac{\beta}{\gamma}+\frac{\gamma}{\alpha}\right)}=-\left(\frac{\alpha}{\gamma}+\frac{\gamma}{\beta}+\frac{\beta}{\alpha}\right)=-c$$This also implies that $|b|=|c|$. We just need to minimise $\frac{4|b|^2}{a}$. We know that $$|b|^2=b\overline{b}=\left(\frac{\alpha}{\beta}+\frac{\beta}{\gamma}+\frac{\gamma}{\alpha}\right)\left(\frac{\alpha}{\gamma}+\frac{\gamma}{\beta}+\frac{\beta}{\alpha}\right)=3+\sum_{cyc}\frac{\alpha\beta}{\gamma}+\sum_{cyc}\frac{\gamma}{\alpha\beta}=3+\sum_{cyc}\alpha^3\beta^3-\sum_{cyc}\alpha^3$$ We also have $\alpha^3+a\alpha+1=0$, and $\alpha+\beta+\gamma=0$. Thus, we can compute $\sum_{cyc}\alpha^3=-3$. We also get that $a=-\frac{\alpha^3+1}{\alpha}$, and since $a$ is a positive real number, this means $\alpha\in(-1,0)$. Also, $$\sum_{cyc}\alpha^3\beta^3=\alpha^3(\beta^3+\gamma^3)+\beta^3\gamma^3=\alpha^3(-3-\alpha^3)-\frac{1}{\alpha^3}$$ Thus, $$\left(\frac{|b|+|c|}{a}\right)^2=\frac{4|b|^2}{a^2}=\frac{4\left(6+\alpha^3(-3-\alpha^3)-\frac{1}{\alpha^3}\right)}{\left(\frac{\alpha^3+1}{\alpha}\right)^2}=\frac{4(-\alpha^9-3\alpha^6+6\alpha^3-1)}{\alpha(\alpha^3+1)^2}$$ Nice work! It appears your minimisation is correct, although the equality case is nasty, at about $x ~ -0.36328$, which gives the exact minimum of $3888^\frac{1}{6}$.

Back to the problem of minimising $\frac{4(-\alpha^9-3\alpha^6+6\alpha^3-1)}{\alpha(\alpha^3+1)^2}$ for $\alpha\in(-1,0)$. Differentiating the function $f(x)=\frac{(-x^9-3x^6+6x^3-1)}{x(x^3+1)^2}$, we have $$f'(x)=-\frac{2x^{12}+5x^9+39x^6-19x^3-1}{x^2(x^3+1)^2}$$which can be set to zero. We let $y=x^3$ for simplicity. $$2y^4+5y^3+39y^2-19y-1=0$$We see that $y=\frac{1}{2}$ is a solution. After long dividing $2y-1$, we obtain the polynomial $y^3+3y^2+21y+1=0$. That is to say, $\alpha^9+3\alpha^6+21\alpha^3+1=0$. This means that $$\frac{4(-\alpha^9-3\alpha^6+6\alpha^3-1)}{\alpha(\alpha^3+1)^2}=\frac{108\alpha^2}{(\alpha^3+1)^2}$$ It now suffices to find the value of $\frac{\alpha^3+1}{\alpha}$. Now after expanding $(\alpha^3+1)^3$, this has a very similar form to the earlier polynomial. In fact, $(\alpha^3+1)^3=-18\alpha^3$, which means that $$\left(\frac{\alpha^3+1}{\alpha}\right)^3=-18$$We can then compute the answer. hsiangshen wrote: Official ans:$3888^{1/6}$ Is there an official solution?

ACGNmath wrote: Is there an official solution? Doesn't seem like it. A quick search (in chinese) shows that only numeric answers are given. That being said, nice job! This feels like a method that would deter many people from trying though. Hopefully someone else has a nicer attempt at this

To summarise my solution, I feel that there are three main ideas: 1) Notice that exactly one root is real, and that the other two roots are complex conjugates of each other (this was what I noticed after assuming that all of $\alpha, \beta, \gamma$ are real and getting a contradiction on the roots). 2) Notice that $b$ and $c$ are complex conjugates of the negative of each other (then facepalm because of course that's what the modulus is there for) 3) Minimise $\frac{4bc}{a^2}$ I have a feeling that if we were to apply some known inequalities like AM-GM or Cauchy, the expressions might be quite contrived, because of the weird equality case. But if anyone has a solution like this, some motivation would be much appreciated. Btw, $a=\sqrt[3]{18}$ is the equality case. If my computations are correct, $b$ and $c$ are $\pm \frac{3}{2}-\frac{3}{2}\sqrt{11}i$.

We can easily get $\alpha\beta\gamma=-1, \alpha+\beta+\gamma=0$ and $\alpha\beta+\beta\gamma+\gamma\alpha=a$ by Vieta's theorem. $\alpha,\beta,\gamma$ are roots of function $x^3+ax+1=0$, so we can get $$\alpha^3+\beta^3+\gamma^3=-a(\alpha+\beta+\gamma)-3=-3$$ Consider the roots of function $x^3+ax^2+1=0$ are $\displaystyle{\frac{1}{\alpha},\frac{1}{\beta},\frac{1}{\gamma}}$ Similarly, $$\displaystyle{(\frac{1}{\alpha})^3+(\frac{1}{\beta})^3+(\frac{1}{\gamma})^3=a^3+3}$$ Lemma:$-bc=a^3+9$ Proof:we can use $\alpha,\beta,\gamma$ to represent $b,c$ (by Vieta's theorem) $\Rightarrow -bc=3-\alpha^3-\beta^3-\gamma^3+\displaystyle{(\frac{1}{\alpha})^3+(\frac{1}{\beta})^3+(\frac{1}{\gamma})^3}=a^3+9$ It's easy to get $-b+c=3$ ,so $-b,c$ are roots of funtion $x^2-3x+a^3+9$ . This funtion implies that $|b|=|c|$ By lemma and AM-GM, $\dfrac{|b|+|c|}{a}=\dfrac{2\sqrt{|bc|}}{a}=\dfrac{2\sqrt{a^3+9}}{a}\geq\dfrac{2\sqrt{3^3\sqrt{3^22^{-2}a^6}}}{a}=2^{\frac{2}{3}}3^{\frac{5}{6}}=3888^{\frac{1}{6}}$