9. (Austrian Mathematical Olympiad 2005) Determine all triples of positive integers $(a, b, c)$ such that $a+b+c$ is the least common multiple of $a, b, c$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
9. Solution: If the three numbers are equal, then we have $[a, b, c]=a$, which contradicts $[a, b, c]=a+b+c$.

Without loss of generality, assume $a \leqslant b \leqslant c$, and the equality does not hold simultaneously, then $a+b<2 c$.
Therefore, $c<a+b+c<3 c$.
Since $[a, b, c]$ is a multiple of $c$, we have $[a, b, c]=2 c$.
Thus, $a+b=c, [a, b, c]=2 a+2 b$,
and since $b \mid(2 a+2 b)$, it follows that $b \mid 2 a$.
From $a \leqslant b$, we can deduce that $b=a$ or $b=2 a$.
If $b=a$, then $c=a+b=2 a$, so $[a, b, c]=[a, a, 2 a]=2 a$.
But $a+b+c=a+a+2 a=4 a$, which is a contradiction.
If $b=2 a$, then $c=a+b=a+2 a=3 a$.
$[a, b, c]=[a, 2 a, 3 a]=6 a=a+b+c$.
Therefore, the form of the three positive integers that satisfy the problem's requirements is $(a, 2 a, 3 a)(a \geqslant 1)$.