Example 1.27 Given a cubic equation $x^{3} + a x^{2} + b x + c = 0 (a, b, c \in \mathbf{R})$ with three roots $\alpha$, $\beta$, $\gamma$ whose magnitudes are all no greater than 1, find
$$\frac{1+|a|+|b|+|c|}{|\alpha|+|\beta|+|\gamma|}$$

the minimum value.
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 $\frac{\sqrt[3]{2}}{2}$.