7. Given complex numbers $z_{1}, z_{2}, z_{3}$ satisfy $\left|z_{1}\right| \leqslant 1,\left|z_{2}\right| \leqslant 1,\left|2 z_{3}-\left(z_{1}+z_{2}\right)\right| \leqslant\left|z_{1}-z_{2}\right|$. Then the maximum value of $\left|z_{3}\right|$ is $\qquad$ .
7. $\sqrt{2}$.

Notice that
$$
\left|2 z_{3}\right|-\left|z_{1}+z_{2}\right| \leqslant\left|2 z_{3}-\left(z_{1}+z_{2}\right)\right| \leqslant\left|z_{1}-z_{2}\right| \text {. }
$$

Then
$$
2\left|z_{3}\right| \leqslant\left|z_{1}+z_{2}\right|+\left|z_{1}-z_{2}\right| \leqslant \sqrt{2\left(\left|z_{1}+z_{2}\right|^{2}+\left|z_{1}-z_{2}\right|^{2}\right)}=\sqrt{4\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)} \leqslant 2 \sqrt{2} \text {. }
$$

When $z_{2}= \pm \mathrm{i} \cdot z_{1}\left(\left|z_{1}\right|=1\right), z_{3}=z_{1}+z_{2}$, $\left|z_{3}\right|$ reaches its maximum value $\sqrt{2}$.