1. Let $P$ be a point in the plane of $\triangle A B C$, the minimum value of $\frac{P A}{P C}+\frac{P B}{A C}+\frac{P C}{A B}$ is
1. $\sqrt{3}$

Let $P, A, B, C$ correspond to the complex numbers $z, z_{1}, z_{2}, z_{3}$ respectively.
Let 
$$
f(z)=\frac{\left(z-z_{1}\right)\left(z-z_{2}\right)}{\left(z_{3}-z_{1}\right)\left(z_{3}-z_{2}\right)}+\frac{\left(z-z_{1}\right)\left(z-z_{3}\right)}{\left(z_{2}-z_{1}\right)\left(z_{2}-z_{1}\right)}+
\frac{\left(z-z_{2}\right)\left(z-z_{3}\right)}{\left(z_{1}-z_{2}\right)\left(z_{1}-z_{3}\right)}
$$

Since $f(z)$ is a quadratic polynomial in $z$, and $f\left(z_{1}\right)=f\left(z_{2}\right)=f\left(z_{3}\right)=1$.
Therefore, $f(z) \equiv 1$.
$$
\begin{array}{l}
\text { Thus } \frac{P A}{B C}+\frac{P B}{A C}+\frac{P C}{A B} \\
\geqslant \sqrt{3\left(\frac{P A \cdot P B}{B C \cdot A C}+\frac{P A \cdot P C}{B C \cdot A B}+\frac{P B \cdot P C}{A C \cdot A B}\right)} \\
=\sqrt{3} \text {. } \\
\sqrt{\left(\frac{\left(z-z_{1}\right)\left(z-z_{2}\right)}{\left(z_{3}-z_{1}\right)\left(z_{3}-z_{2}\right)}\right)+\left|\frac{\left(z-z_{1}\right)\left(z-z_{3}\right)}{\left(z_{2}-z_{1}\right)\left(z_{2}-z_{3}\right)}\right|+\left|\frac{\left(z-z_{2}\right)\left(z-z_{3}\right)}{\left(z_{1}-z_{2}\right)\left(z_{1}-z_{3}\right)}\right|} \\
\geqslant \sqrt{3} \cdot 1=\sqrt{3} \text {. } \\
\end{array}
$$