1. In $\triangle A B C$, $A D \perp B C$ at $D$, and $\frac{A D}{B C}=\frac{1}{3}$. Then the maximum value of $\frac{A C}{A B}+\frac{A B}{A C}$ is . $\qquad$
$=\sqrt{1} \cdot \sqrt{13}$.
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 $\sin (A+\varphi)=1$, $\left(\frac{A C}{A B}+\frac{A B}{A C}\right)_{\max }=\sqrt{13}$.