3. Let the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>1, b>1)$ have a focal distance of $2 c$, and let the line $l$ pass through the points $(a, 0)$ and $(0, b)$. If the distance from the point $(1, 0)$ to the line $l$ plus the distance from the point $(-1,0)$ to the line $l$ is $S \geqslant \frac{4 c}{5}$. Then the range of the eccentricity $e$ of the hyperbola is
3. $\left[\frac{\sqrt{5}}{2}, \sqrt{5}\right]$.

Let the line $l: \frac{x}{a}+\frac{y}{b}=1$.
Then, the distance from the point $(1,0)$ to the line $l$ is
$$
\frac{\left|\frac{1}{a}-1\right|}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}}=\frac{|b-a b|}{c}=\frac{(a-1) b}{c}(a>1),
$$

The distance from the point $(-1,0)$ to the line $l$ is
$$
\frac{\left|-\frac{1}{a}-1\right|}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}}=\frac{(a+1) b}{c} \text {. }
$$

Then, $S=\frac{(a+1) b}{c}+\frac{(a-1) b}{c}=\frac{2 a b}{c} \geqslant \frac{4}{5} c$
$\Leftrightarrow 5 a b \geqslant 2 c^{2} \Leftrightarrow 5 a \sqrt{c^{2}-a^{2}} \geqslant 2 c^{2}$
$\Leftrightarrow 5 \sqrt{e^{2}-1} \geqslant 2 e^{2} \Leftrightarrow 4 e^{4}-25 e^{2}+25 \leqslant 0$
$\Leftrightarrow\left(4 e^{2}-5\right)\left(e^{2}-5\right) \leqslant 0 \Leftrightarrow \frac{5}{4} \leqslant e^{2} \leqslant 5$.
Since $e>1$, thus, $\frac{\sqrt{5}}{2} \leqslant e \leqslant \sqrt{5}$.