3. In the complex plane, the points $0, z, \frac{1}{z}, z+\frac{1}{z}$ form a parallelogram with an area of $\frac{4}{5}$. Then the minimum value of $\left|z+\frac{1}{z}\right|$ is
3. $\frac{2 \sqrt{5}}{5}$.

It is known that, $\frac{1}{z}=\frac{\bar{z}}{|z|^{2}}$. Therefore, the arguments of $\frac{1}{z}$ and $z$ on the complex plane are opposite numbers.
Assume $z=r \mathrm{e}^{\mathrm{i} \theta}$. Then
$$
\frac{1}{z}=\frac{\mathrm{e}^{-\mathrm{i} \theta}}{r} \text {. }
$$

Thus, the area of the parallelogram can be expressed as
$$
\begin{array}{l}
|z|\left|\frac{1}{z}\right| \sin 2 \theta \\
\Rightarrow \sin 2 \theta=\frac{4}{5} . \\
\text { Hence }\left|z+\frac{1}{z}\right|^{2}=\left(r \mathrm{e}^{\mathrm{i} \theta}+\frac{\mathrm{e}^{-\mathrm{i} \theta}}{r}\right)\left(r \mathrm{e}^{-\mathrm{i} \theta}+\frac{\mathrm{e}^{\mathrm{i} \theta}}{r}\right) \\
=r^{2}+\mathrm{e}^{\mathrm{i} 2 \theta}+\mathrm{e}^{-\mathrm{i} 2 \theta}+\frac{1}{r^{2}} \\
=r^{2}+\frac{1}{r^{2}}+2 \cos 2 \theta \geqslant \frac{4}{5} .
\end{array}
$$

Therefore, $\left|z+\frac{1}{z}\right| \geqslant \frac{2 \sqrt{5}}{5}$.