1. Given three positive real numbers $a$, $b$, $c$ satisfying $a+b+c=1$. Then the minimum value of $\frac{1-a b-b c-c a}{\sqrt{a b c}}$ is $\qquad$
$1.2 \sqrt{3}$.
$$
\begin{array}{l}
\text { Since } a b+b c+c a \leqslant \frac{(a+b+c)^{2}}{3}=\frac{1}{3}, \\
\sqrt{a b c} \leqslant \sqrt{\left(\frac{a+b+c}{3}\right)^{3}}=\frac{\sqrt{3}}{9},
\end{array}
$$

Therefore, the minimum value of $\frac{1-a b-b c-c a}{\sqrt{a b c}}$ is $2 \sqrt{3}$.