Given the problem, let's assume $1 \geqslant|\alpha| \geqslant|\beta| \geqslant|\gamma|, \beta=s \alpha, \gamma=t \alpha$, then $1 \geqslant|s| \geqslant|t|$, and let

then
$$\begin{aligned} \theta= & |\alpha| \leqslant 1, u=\frac{1+|a|+|b|+|c|}{|\alpha|+|\beta|+|\gamma|} \\ u= & \frac{1+|a|+|b|+|c|}{|\alpha|+|\beta|+|\gamma|}= \\ & \frac{1+\theta|1+s+t|+\theta^{2}|s+t+s t|+\theta^{3}|s t|}{\theta(1+|s|+|t|)} \geqslant \\ & \frac{1+\theta^{3}(|1+s+t|+|s+t+s t|+|s t|)}{3 \theta} \geqslant \\ & \frac{1+\theta^{3}(|1+s+t|+|s+t|)}{3 \theta} \geqslant \frac{1+\theta^{3}}{3 \theta}= \\ & \frac{1}{3}\left(\frac{1}{2 \theta}+\frac{1}{2 \theta}+\theta^{2}\right) \geqslant \sqrt[3]{\frac{1}{2 \theta} \cdot \frac{1}{2 \theta} \cdot \theta^{2}}=\frac{\sqrt[3]{2}}{2} \end{aligned}$$

It is easy to verify that when $\theta=\frac{\sqrt[3]{4}}{2}, s=-\frac{-1+\sqrt{3} \mathrm{i}}{2}, t=-\frac{-1-\sqrt{3} \mathrm{i}}{2}$, $u$ reaches its minimum value $\frac{\sqrt[3]{2}}{2}$.