1. Given positive real numbers $a$, $b$, $c$ satisfy
$$
(1+a)(1+b)(1+c)=8 \text {. }
$$

Then the minimum value of $a b c+\frac{9}{a b c}$ is
-1.10 .
$$
\begin{array}{l}
\text { Given } 8=(1+a)(1+b)(1+c) \\
\geqslant 2 \sqrt{a} \cdot 2 \sqrt{b} \cdot 2 \sqrt{c}=8 \sqrt{a b c} \\
\Rightarrow a b c \leqslant 1 .
\end{array}
$$

Equality holds if and only if $a=b=c=1$.
It is easy to verify that the function
$$
f(x)=x+\frac{9}{x}
$$

is decreasing on $(0,3)$.
Since $0<a b c \leqslant 1<3$, when $a b c=1$,
$$
f(a b c)=a b c+\frac{9}{a b c}
$$
is minimized, and the minimum value is 10.