7. For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>1, b>0)$, the focal distance is $2 c$. 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 $\qquad$
For the line $l: b x+a y-a b=0$, then $s=\frac{|b-a b|}{c}+\frac{|-b-a b|}{c}=\frac{2 a b}{c}$, thus $\frac{2 a b}{c} \geqslant \frac{4}{5} c \Rightarrow 5 a b \geqslant 2 c^{2} \Rightarrow 2 b^{2}-5 a b+2 a^{2} \leqslant 0 \Rightarrow \frac{1}{2} \leqslant \frac{b}{a} \leqslant 2$, so $e^{2}=1+\frac{b^{2}}{a^{2}} \in\left[\frac{5}{4}, 5\right] \Rightarrow e \in\left[\frac{\sqrt{5}}{2}, \sqrt{5}\right]$.