Proving At Least One Real Number Is Greater Than Average

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To prove that at least one of the real numbers a1, a2, a3, ..., an is greater than or equal to their average, consider that the average is defined as the total sum divided by the count of numbers. If all numbers were less than the average, their total would be less than the average multiplied by the count, which is a contradiction. For part (b), using the integers 1 through 10 arranged in a circle, calculate the sums of all consecutive triplets. Since the total sum of these integers is 55, the average is 5.5, making it necessary for at least one triplet to exceed a sum of 17. This reasoning confirms that at least one triplet sum must be greater than or equal to 17.
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a said:
Show that at least one of the real numbers a1, a2, a3, ..., an is greater or equal to the average of these numbers.
b said:
Use the result in (a) to show that if the first 10 positive integers are placed around a circle, in any order, there exist three integers in consecutive locations around the circle that have a sum greater than or equal to 17
i can solve question (a) but not (b)
i totally don't know where to start, can anyone give me a some hints?
 
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Maybe by contradiction?
 
Let's name the integers around the circle as
a_1, a_2, ..., a_10 (they are the integers 1 through 10 in some order).

Now let's go around the circle calculating partial sums
s_1 = a_1 + a_2 + a_3
s_2 = a_2 + a_3 + a_4
...
s_8 = a_8 + a_9 + a_10
s_9 = a_9 + a_10 + a_1
s_10= a_10 + a_1 + a_2
All possible triplets of integers in consecutive locations around the circle are represented here, as well as their sums.
We can rephrase the question now:
prove that there is at least one of those sums greater than or equal to 17.

Does that help?
 
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help!

i was hoping somebody could show me how to solve this problem...what proof would you use?
 
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