Consider the complex numbers $z_{1}, z_{2}, z_{3}, z_{4}, z_{5}$ satisfying:
$$
\left\{\begin{array}{l}
\left|z_{1}\right| \leqslant 1,\left|z_{2}\right| \leqslant 1 \\
\left|2 z_{3}-\left(z_{1}+z_{2}\right)\right| \leqslant\left|z_{1}-z_{2}\right| \\
\left|2 z_{4}-\left(z_{1}+z_{2}\right)\right| \leqslant\left|z_{1}-z_{2}\right| \\
\left|2 z_{5}-\left(z_{3}+z_{4}\right)\right| \leqslant\left|z_{3}-z_{4}\right|
\end{array},\right.
$$

Find the maximum value of $\left|z_{5}\right|$.
Given the known inequality and $\left|z_{1}-z_{2}\right|+\left|z_{1}+z_{2}\right| \leqslant 2 \sqrt{\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}}$, we have
4
$$
\begin{aligned}
\left|z_{5}\right|= & 2\left|2 z_{5}-\left(z_{3}+z_{4}\right)+\left(z_{3}+z_{4}\right)\right| \\
\leqslant & 2\left[\left|2 z_{5}-\left(z_{3}+z_{4}\right)\right|+\left|z_{3}+z_{4}\right|\right] \\
\leqslant & 2\left(\left|z_{3}-z_{4}\right|+\left|z_{3}+z_{4}\right|\right) \\
= & \left|\left[2 z_{3}-\left(z_{1}+z_{2}\right)\right]-\left[2 z_{4}-\left(z_{1}+z_{2}\right)\right]\right| \\
& +\left|\left[2 z_{3}-\left(z_{1}+z_{2}\right)\right]+\left[2 z_{4}-\left(z_{1}+z_{2}\right)\right]+2\left(z_{1}+z_{2}\right)\right| \\
\leqslant & \left|\left[2 z_{3}-\left(z_{1}+z_{2}\right)\right]-\left[2 z_{4}-\left(z_{1}+z_{2}\right)\right]\right| \\
& +\left|\left[2 z_{3}-\left(z_{1}+z_{2}\right)\right]+\left[2 z_{4}-\left(z_{1}+z_{2}\right)\right]\right|+2\left|z_{1}+z_{2}\right| \\
\leqslant & 2\left|z_{1}+z_{2}\right|+2 \sqrt{\left|2 z_{3}-\left(z_{1}+z_{2}\right)\right|^{2}+\left|2 z_{4}-\left(z_{1}+z_{2}\right)\right|^{2}}
\end{aligned}
$$
$$
\begin{array}{l}
\leqslant 2\left|z_{1}+z_{2}\right|+2 \sqrt{2}\left|z_{1}-z_{2}\right| \\
\leqslant 2 \sqrt{\left[1^{2}+(\sqrt{2})^{2}\right] \cdot\left(\left|z_{1}+z_{2}\right|^{2}+\left|z_{1}-z_{2}\right|^{2}\right)} \\
=2 \sqrt{3} \cdot \sqrt{2\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)} \\
\leqslant 2 \sqrt{3} \cdot \sqrt{4}=4 \sqrt{3} .
\end{array}
$$

Therefore, $\left|z_{5}\right| \leqslant \sqrt{3}$. When $z_{1}=e^{i \theta}, z_{2}=e^{-i \theta}, z_{3}=\frac{1}{2}\left(z_{1}+z_{2}\right)+\frac{\sqrt{6}}{3} e^{\frac{\pi}{4}}, z_{4}=\frac{1}{2}\left(z_{1}+z_{2}\right)+$ $\frac{\sqrt{6}}{3} e^{-i \frac{\pi}{4}}, z_{5}=\sqrt{3}$, the equality holds, where $\theta=\arctan \sqrt{2}$. Hence, $\left|z_{5}\right|_{\max }=\sqrt{3}$.