5. In $\triangle A B C$, the lengths of the three sides $a, b, c$ satisfy $b+c \leqslant 2 a$ and $a+c \leqslant 2 b$. Then the range of $\frac{b}{a}$ is $\qquad$
5. $\frac{2}{3}a-b$, given $2 a-b<a+c<2 b$, i.e., $\frac{2}{3}<\frac{b}{a} \leqslant 1$.
When $a \leqslant b$, we have $1 \leqslant \frac{b}{a}<\frac{3}{2}$. Therefore, $\frac{2}{3}<\frac{b}{a}<\frac{3}{2}$.