There is a third possible interpretation of the question - the largest possible score with particular subscores that guarantees the finals. This is the highest possible score that gets a candidate to the finals no matter how others arrange their scores. But it isn't just the score that guarantees the finals - subscores need to be just right.
As an example, subscores 2,2,20 with total score of 80 guarantees us the spot.
The first obvious observation - we assign some scores to our candidate. Then, 8 competitors want to be tied or ahead (to possibly cut candidate out of the finals), and we assign the lowest remaining numbers to those 8. It is obvious that there is no reason to pick any higher number - so picking numbers 1-9 is sufficient for these candidates.
To simplify some math, trivial observation is that you are interested in geometric mean instead of the final score. Final score is a cube of geometric mean of individual scores.
So, these mentioned subscores have a geomean of 4.31. Remaining subscores to be assigned (3x1-8 except 2, 2x9, 1x 2) have a geomean of 4.27. Uh oh, trouble. However, let's try to assign these scores to people. Obviously, we pair 1 with 8 and 9 twice, then pair last 1 with 8 and 7 - say 189, 918, 871. Now, the remaining scores have geomean of 4.41 = we are on the safe side because geometric mean of remaining scores is at least 85 and something, so not all can be ahead of our candidate.
Now, what is the highest possible score where that is possible?
We have 3 triplets - 189, 257, 346 with scores 72, 70, 72. We need to break them up somehow to squeeze in some higher number. There are two options - we remove 1 subscore 1-3 or we remove 2. (it is plainly obvious that removal of just say 2x 4 simply means one final triplet will be 336 and the other 346)
Let's first check the easier case:
Removal of one subscore. If we remove 2 digits from same triplet, we didn't do anything useful, so we need to remove from 2 different triplets. But if we say removed 3 and 5, one of that 5 is replaced by 4 present in 346 triplet we eliminated. There are 2 reasonable cases - 24x and 1 + 5-7.
Let's check that out
Removal of 7 is trivial, replace with 8 for score 80, so we are at best matching already achieved score. Removal of 6 requires 347 (and 258) with score 84. But 1,6,13 is 78 and 1,6,14 is same 84, so this doesn't work. Finally, removal of 5 offers scores of either 80 and 85 and it is easy to arrange for score 84 (189x2, 258, 257, 266, 346, 347x2).
What remains are possibilities 23x, 24x and 33x. 24x is not going to work, it is possible to get 8 scores under or equal to 84, so 88 is out of the question while 80 has been achieved already. 23x offers 84, but other combinations 189 x3 + 267 x2 + 355 + 445 + 346 sink the option. Finally, 339 offers score 81. To remove 1s, we use 189 x3, for removal of 2 we take 257 x2 and 266 and we are left with 3, 3x4, 5, 6; which we arrange to 346, 445, which score 72 and 80. So, 339 is also impossible.
Therefore, the first 2,2,20 is the highest score of 80 that guarantees the spot in the finals.
(there might be other combinations with the same score that also guarantee spot in the finals - but it is easy to see that 445 does not - 189x3, 266, 257x2, 346, 337 are all lower)