How find this inequality find the maximum $z_{5}$ let $z_{1},z_{2},z_{3},z_{4},z_{5}\in C$,such

$$\begin{cases}
|z_{1}|\le 1,|z_{2}|\le 1\\
|2z_{3}-(z_{1}+z_{2})|\le|z_{1}-z_{2}|\\
|2z_{4}-(z_{1}+z_{2})|\le|z_{1}-z_{2}|\\
|2z_{5}-(z_{3}+z_{4})|\le|z_{3}-z_{4}|
\end{cases}$$
  Find the maximum $|z_{5}|$

My ugly solution:
note 
$$|z_{1}-z_{2}|+|z_{1}+z_{2}|\le 2\sqrt{|z1_{1}|^2+|z_{2}|^2}$$
then we have

\begin{align*}&4|z_{5}|=2|2z_{5}-(z_{3}+z_{4})+(z_{3}+z_{4})|\le 2[|2z_{5}-(z_{3}+z_{4})|+|z_{3}+z_{4}|]\\
&\le 2(|z_{3}-z_{4}|+|z_{3}+z_{4}|)\\
&=|[2z_{3}-(z_{1}+z_{2})]-[2_{4}-(z_{1}+z_{2})]|+|[2z_{3}-(z_{1}+z_{2})]+[2z_{4}-(z_{1}+z_{2})]+2(z_{1}+z_{2})|\\
&\le|[2z_{3}-(z_{1}+z_{2})]-[2_{4}-(z_{1}+z_{2})]|+|[2z_{3}-(z_{1}+z_{2})]+[2z_{4}-(z_{1}+z_{2})]|+2|(z_{1}+z_{2})|\\
&\le 2|z_{1}+z_{2}|+2\sqrt{|2z_{3}-(z_{1}+z_{2})|^2+|2z_{4}-(z_{1}+z_{2})|^2}\\
&\le 2|z_{1}+z_{2}|+2\sqrt{2}|z_{1}-z_{2}|\\
&\le2\sqrt{[1^2+(\sqrt{2})^2]\cdot(|z_{1}+z_{2}|^2+|z_{1}-z_{2}|^2)}\\
&=2\sqrt{3}\cdot\sqrt{2(|z_{1}|^2+|z_{2}|^2)}\\
&\le2\sqrt{3}\cdot \sqrt{4}=4\sqrt{3}
\end{align*}
  so
  $$|z_{5}|\le \sqrt{3}$$
  if and only if $z_{1}=e^{i\theta},z_{2}=e^{-\theta},z_{3}=\dfrac{1}{2}(z_{1}+z_{2})+\dfrac{\sqrt{6}}{3}e^{\frac{\pi}{4}i},z_{4}=\dfrac{1}{2}(z_{1}+z_{2})+\dfrac{\sqrt{6}}{3}e^{-\frac{\pi}{4}i},z_{5}=\sqrt{3},\theta=\arctan{\sqrt{2}}$

My question: Have other nice methods? or can someone can use geometric Explain and solve the problem?

It seems the following.
Your solution is nice but I shall use a geometric approach. 
The points $z_1$ and $z_2$ belong to a disk $D_1$ of radius $R_1=1$ centered at the point $o_1=0$. The points $z_3$ and $z_4$ belong to a disk $D_2$ of radius $R_2=|z_1-z_2|/2$ centered at the point $o_2=(z_1+z_2)/2$. The point $z_5$ belongs to a disk of radius $R_3=|z_3-z_4|/2$ centered at the point $o_3=(z_3+z_4)/2$. Put $r_1=|o_1o_2|$ and $r_2=|o_2o_3|$. Since $z_1,z_2\in D_1$, then $$R_2^2=|z_1o_2||z_2o_2|\le (R_1+r_1)(R_1–r_1)=R_1^2-r_1^2$$ (here we use the theorem claiming that the product of  lengths of the parts of the chords going through a fixed point is constant). Similarly, $$R_3^2=|z_3o_3||z_3o_4|\le (R_2+r_2)(R_2–r_2)=R_2^2-r_2^2=1-r_1^2-r_2^2.$$ Moreover, $|O_3Z_5|\le R_3$. Now $$|z_5|=|O_1Z_5|\le |o_1o_2|+|o_2o_3|+|o_3z_5|=r_1+r_2+\sqrt{1-r_1^2-r_2^2}\le$$ $$3\sqrt{\frac {r_1^2+r^2_2+1- r_1^2-r_2^2}{3}}=\sqrt{3},$$
and the equality is possible only if the points $o_2$ and $o_3$ consequently lie on a segment $[o_1z_5]$ and $r_1=r_2=\frac 1{\sqrt 3}$. 
