5. Given positive numbers $a, b, x, y, z$ satisfying $a \leqslant x, y, z \leqslant b$, and the difference between the maximum and minimum values of $(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ is $\frac{9}{5}$, then $\frac{2 b}{a}=$ $\qquad$ .
$5$