5. Given that the three sides $a, b, c$ of $\triangle A B C$ satisfy:
$$
a+b \leqslant 2 c, b+c \leqslant 3 a \text {. }
$$

Then the range of $\frac{c}{a}$ is $\qquad$
5. $\left(\frac{2}{3}, 2\right)$.

From the problem, we know that $a+b+c \leqslant 3 c, a+b+c \leqslant 4 a$.
Let $s=a+b+c=1$. Then $c \geqslant \frac{1}{3}, a \geqslant \frac{1}{4}, a+c>\frac{1}{2}$.
If $c>a \Rightarrow \max \left\{\frac{1}{3}, a\right\} \leqslant c \leqslant \frac{1}{2}$,
If $c<a \Rightarrow \frac{1}{3} \leqslant c \leqslant \max \left\{\frac{1}{2}, a\right\}$,
If $c=a \Rightarrow \frac{c}{a}=1$.
In summary, $\frac{c}{a} \in\left(\frac{2}{3}, 2\right)$.