For any arrangment of numbers 1 to 10 in a circle, there will always exist a pair of 3 adjacent numbers in the circle that sum up to 17 or more I set out to solve the following question using the pigeonhole principle

Regardless of how one arranges numbers $1$ to $10$ in a circle, there will always exist a pair of three adjacent numbers in the circle that sum up to $17$ or more.

My outline
[1] There are $10$ triplets consisting of adjacent numbers in the circle, and since each number appears thrice, the total sum of these adjacent triplets for all permutations of the number in the circle, is $3\cdot 55=165$.
[2] If we consider that all the adjacent triplets sum to 16 , and since there are $10$ such triplets, the sum accordingly would be $160$, but we just said the invariant sum is $165$ hence there would have to be a triplet with sum of $17$ or more.
My query
Could someone polish this into a mathematical proof and also clarify if I did make use of the pigeonhole principle.   

We will show something stronger, namely that there exists 3 adjacent numbers that sum to 18 or more.
Let the integers be $\{a_i\}_{i=1}^{10}$. WLOG, $a_1 = 1$. Consider
$$a_2 + a_3 + a_4, a_5 + a_6 + a_7, a_8 + a_9 + a_{10}$$
The sum of these 3 numbers is $2+3 +\ldots + 10 = 54$. Hence, by the pigeonhole principle, there exists one number which is at least $\lfloor \frac{54}{3} \rfloor = 18 $.

I leave it to you to show that there is a construction where no 3 adjacent numbers sum to 19 or more, which shows that 18 is the best that we can do.
