If $\alpha, \beta, \gamma$ are the roots of $x^3 + (a^4 + 4a^2 + 1)x = x^2 + a^2$ then minimum value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} + \frac{\gamma}{\alpha} + \frac{\alpha}{\gamma} + \frac{\beta}{\gamma} + \frac{\gamma}{\beta}$ is?
I read this questions and tried to attempt it. The answer to this question on solving comes out to be $3$. However the part where I am having trouble understanding is that why is $AM \geq GM$ not valid in this case? Since the GM of the expression comes out to be 1, therefore the inequality tells me that the given expression $\geq 6$ which is clearly not the case here!
As the comments have already pointed out, The AM-GM inequality is only valid for positive real numbers, which the roots very well might not be. Nevertheless, here's a proof of why the minimum value of the given expression is three (assuming $ a \in \mathbb{R} $ and $a \neq 0 $).
Since we have
$ \displaystyle \begin{align} \frac{\alpha}{\beta}+\frac{\alpha}{\gamma}+\frac{\gamma}{\beta}+ \frac{\gamma}{\alpha}+ \frac{\beta}{\gamma}+\frac{\beta}{\alpha} &= \frac{ \alpha^2(\gamma +\beta) +\beta^2(\alpha+\gamma)+\gamma^2(\beta+\alpha)}{\alpha \beta \gamma} \\ &= \frac{\alpha^2 +\beta^2 + \gamma^2-(\alpha^3+\beta^3+\gamma^3)}{\alpha \beta \gamma} \\ &= \frac{ (\alpha +\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)-(\alpha +\beta+\gamma)((\alpha +\beta+\gamma)^2-3(\alpha\beta+\beta\gamma+\gamma\alpha))-3 \alpha \beta \gamma}{\alpha\beta \gamma} \\ &= \frac{a^4+a^2+1}{a^2} \\ &= 1+a^2+\frac{1}{a^2} \\ &\ge 3 \end{align} \tag*{} $