7. Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>1, b>0)$ with focal distance $2 c$, the line $l$ passes through the points $(a, 0)$ and $(0, b)$, and the sum of the distances from the point $(1,0)$ to the line $l$ and from the point $(-1,0)$ to the line $l$ is $s \geqslant \frac{4}{5} c$. Then the range of the eccentricity $e$ of the hyperbola is $\qquad$.
7. $\left[\frac{\sqrt{5}}{2}, \sqrt{5}\right]$.

Let the line $l: \frac{x}{a}+\frac{y}{b}=1$, i.e.,
$$
b x+a y-a b=0 \text {. }
$$

By the distance formula from a point to a line, and given $a>1$, the distance $d_{1}$ from the point $(1,0)$ to the line $l$ is $\frac{b(a-1)}{\sqrt{a^{2}+b^{2}}}$; the distance $d_{2}$ from the point $(-1,0)$ to the line $l$ is $\frac{b(a+1)}{\sqrt{a^{2}+b^{2}}}$. Thus, $s=d_{1}+d_{2}=\frac{2 a b}{\sqrt{a^{2}+b^{2}}}=\frac{2 a b}{c}$.
$$
\begin{array}{l}
\text { By } s \geqslant \frac{4}{5} c \Rightarrow 5 a \sqrt{c^{2}-a^{2}} \geqslant 2 c^{2} \\
\Rightarrow 5 \sqrt{e^{2}-1} \geqslant 2 e^{2} \\
\Rightarrow 4 e^{4}-25 e^{2}+25 \leqslant 0 \\
\Rightarrow \frac{5}{4} \leqslant e^{2} \leqslant 5(e>1) \\
\Rightarrow \frac{\sqrt{5}}{2} \leqslant e \leqslant \sqrt{5} .
\end{array}
$$