Problem 615

$=\sqrt{1} \cdot \sqrt{13}$.
Let $A B=c, B C=a, C A=b, A D=h$.
By the cosine rule, we get
$$b^{2}+c^{2}=a^{2}+2 b c \cos A \text {. }$$

By the principle of equal area, we get $a h=b c \sin A$.
$$\begin{array}{l} \because \frac{h}{a}=\frac{1}{3} \Rightarrow a^{2}=3 b c \sin A, \\ \therefore b^{2}+c^{2}=3 b c \sin A+2 b c \cos A \\ \Rightarrow \frac{A C}{A B}+\frac{A B}{A C}=\frac{b}{c}+\frac{c}{b}=3 \sin A+2 \cos A \\ =\sqrt{13} \sin (A+\varphi) . \text { where } \varphi=\arctan \frac{2}{3} . \end{array}$$

Therefore, when $\sin (A+\varphi)=1$, $\left(\frac{A C}{A B}+\frac{A B}{A C}\right)_{\max }=\sqrt{13}$.