Proof from Discrete Mathematics

Question

I am trying to prove a result

If the first $10$ positive integers are placed around a circle, in any order, then there exists three integers in consecutive locations around the circle that have a sum greater than or equal to 17.

I am trying to prove it using contradiction. But I am not getting what will be the negation of the conclusion?

Will it be, for any three integers in consecutive locations, the sum is strictly less than 17? If so, I dont know how to proceed ahead. If not, what will be the correct negation.

The reason for asking this question here is that I really think the negation I am doing is wrong!

Answer 1

Hint:

Let the numbers be in the order $[a_1,a_2,a_3,a_4,\dots,a_9,a_{10}]$ where it wraps around.

Consider the sum $(a_1+a_2+a_3)+(a_2+a_3+a_4)+(a_3+a_4+a_5)+\dots+(a_9+a_{10}+a_1)+(a_{10}+a_1+a_2)$

Answer 2

"Will it be, for any three integers in consecutive locations, the sum is strictly less than 17?"

Yes.

so as a contradiction if you could attempt to show that $(a+b+c) + (d+e+f) + (g + h + i) < 3*17 = 51$.

But $1 + 2 + 3+..... + 10= a+b+c+d+e+f+g+h+i+j = 55$ so $55 = a+b+c+d+e+f+g+h+i+j < 51 + j$ so $4 < j$.

Can you get a contradiction from that?

Hint: How did I choose $j$ to be left out when I declared $(a+b+c) + (d+e+f) + (g + h + i) < 3*17 = 51$? What would have happened if I had chosen, say, $e$?