Proof: There exists a complex number $z_{0}$ satisfying $\left|z_{0}\right|=1$, such that
$$\left|f\left(z_{0}\right)\right| \geqslant \frac{\prod_{i=1}^{n}\left(1+\left|\alpha_{j}\right|\right)}{2^{n-1}}.$$

【Analysis and Proof】First, take a step back and consider a special case.
Assume $\alpha_{j}$ lies on the unit circle. First, prove a lemma.
Lemma 1 Assume $\left|\alpha_{j}\right|=1$. Then there exists a complex number $z_{0}$ satisfying $\left|z_{0}\right|=1$, such that $\left|f\left(z_{0}\right)\right| \geqslant 2$.
Proof of Lemma 1 Let
$$f(z)=z^{n}+c_{n-1} z^{n-1}+\cdots+c_{1} z+c_{0}.$$

Then $c_{0}=(-1)^{n} \alpha_{1} \alpha_{2} \cdots \alpha_{n},\left|c_{0}\right|=1$.
Let $\omega \in \mathbf{C}$, such that $\omega^{n}=c_{0}, \omega_{j}=\mathrm{e}^{\frac{3 i \times i}{n}}$, i.e., $\omega_{j}$ is the $j$-th ($j=0,1, \cdots, n-1$) $n$-th root of unity.
Let $x_{j}=\omega \cdot \omega_{j}$. Then $\left|x_{j}\right|=\left|\omega \cdot \omega_{j}\right|=1$, and
$$\begin{array}{l} \left|f\left(x_{0}\right)\right|+\left|f\left(x_{1}\right)\right|+\cdots+\left|f\left(x_{n-1}\right)\right| \\ \geqslant\left|f\left(x_{0}\right)+f\left(x_{1}\right)+\cdots+f\left(x_{n-1}\right)\right| \\ =\left|\sum_{j=0}^{n-1} f\left(x_{j}\right)\right| \\ =\left|\sum_{j=0}^{n-1}\left(x_{j}^{n}+c_{n-1} x_{j}^{n-1}+\cdots+c_{1} x_{j}+c_{0}\right)\right| \\ =\left|n \omega^{n}+0+\cdots+0+n c_{0}\right| \\ =\left|2 n c_{0}\right|=2 n. \end{array}$$

Therefore, among $\left|f\left(x_{j}\right)\right|(j=0,1, \cdots, n-1)$, there must exist one $\left|f\left(x_{i}\right)\right| \geqslant 2$.
Returning to the original problem.
$$\text { Equation (1) } \Leftrightarrow \frac{\prod_{j=1}^{n}\left|z_{0}-\alpha_{j}\right|}{\prod_{j=1}^{n}\left(1+\left|\alpha_{j}\right|\right)} \geqslant \frac{1}{2^{n-1}}.$$

By Lemma 1, there exists $z_{0}$ satisfying $\left|z_{0}\right|=1$, such that
$$\prod_{j=1}^{n}\left|z_{0}-\beta_{j}\right| \geqslant 2,$$

where $\beta_{j}$ has the same principal argument as $\alpha_{j}$, and $\left|\beta_{j}\right|=1$.
Thus, $\frac{\prod_{j=1}^{n}\left|z_{0}-\beta_{j}\right|}{\prod_{j=1}^{n}\left(1+\left|\beta_{j}\right|\right)} \geqslant \frac{2}{2^{n}}=\frac{1}{2^{n-1}}$.
To prove Equation (2), it suffices to prove the following Lemma 2.
Lemma 2 For any $j$ and a complex number $z_{0}$ on the unit circle, $\frac{\left|z_{0}-\alpha_{j}\right|}{1+\left|\alpha_{j}\right|} \geqslant \frac{\left|z_{0}-\beta_{j}\right|}{1+\left|\beta_{j}\right|}$.

Proof of Lemma 2 Let the points corresponding to the complex numbers $z_{0}, \alpha_{j}, \beta_{j}, -\beta_{j}$ in the complex plane be $Z, A, B, B'$, respectively. If point $Z$ lies on $AB$, then $Z$ coincides with point $B$ or $B'$. This is clearly true.

Otherwise, points $Z, A, B'$ form $\triangle ZAB'$ (as shown in Figure 1).

In $\triangle ZAB'$ and right $\triangle ZBB'$, we have
$$\begin{array}{l} \frac{\left|z_{0}-\alpha_{j}\right|}{1+\left|\alpha_{j}\right|}=\frac{ZA}{AB'}=\frac{\sin \angle ZB'A}{\sin \angle B'ZA} \\ \geqslant \sin \angle ZB'B=\frac{ZB}{BB'}=\frac{\left|z_{0}-\beta_{j}\right|}{1+\left|\beta_{j}\right|}. \end{array}$$

In summary, the problem is established.
References: Publication. 1993,3.
[2] Zhu Huachuan, Qian Dianxu, Mathematical Problem Solving Strategies [M]. Beijing: Science Press. 2009.8.