9. Let $a$, $b$, $c$ be the lengths of the sides of $\triangle ABC$, simplify
$$
\sqrt{(a-b-c)^{2}}+\sqrt{(b-c-a)^{2}}+\sqrt{(c-a-b)^{2}}
$$

The result is $\qquad$.
$\begin{array}{l}=9 . a+b+c . \\ \sqrt{(a-b-c)^{2}}+\sqrt{(b-c-a)^{2}}+\sqrt{(c-a-b)^{2}} \\ =(b+c-a)+(c+a-b)+(a+b-c) \\ =a+b+c .\end{array}$

The translation is as follows:

$\begin{array}{l}=9 . a+b+c . \\ \sqrt{(a-b-c)^{2}}+\sqrt{(b-c-a)^{2}}+\sqrt{(c-a-b)^{2}} \\ =(b+c-a)+(c+a-b)+(a+b-c) \\ =a+b+c .\end{array}$