No, you did not prove the negation of the original claim. Your method does prove, however, that there are three consecutive numbers whose sum is at least 17. Indeed, if all three-number sums are at most 16, then their sum is 160, while it has to be 165. Of course, the original claim is stronger.
You are asked to investigate whether a permutation exists such that all three-number sums are at most 17. There are two options: either exhibit such permutation, or show that its existence leads to a contradiction. (See a footnote below.) In the first case, it is not enough to find some plausible facts about such permutation. (In your solution, such fact is that the sum of all three-number sums is <= 170, which is true.) An imaginary object can have some plausible characteristics: for example, a unicorn has four legs. It is also not clear which true facts are enough to establish the existence of the required object. In the limit, I could say, "Suppose that such object exists. But 2 + 2 = 4 is true; therefore, it does, in fact, exists." In reality, of course, nothing shorter than describing the required object in full detail is enough.
One solution can be found on StackExchange.
A small remark: It is not necessary to say that the numbers are scattered at random around a circle. The word "random" may convey the right intuition: that we don't assume anything about the order, but there is nothing related to probability theory about this problem. It is enough to ask to show that the required three-number sum exists for all possible orders (permutations).
Footnote: Strictly speaking, there is a third option: show that the assumption that such permutation does not exist leads to a contradiction. The net result is that the required permutation does exist, just like in the first option. However, first, there are few cases where such proof is shorter or clearer and, second, for finite objects such as permutations of 10 numbers, such proof can be converted into a proof of the first option, which explicitly constructs a permutation.