Integers $1, 2, \ldots, 10$ are circularly arranged in an arbitrary order. 
Possible Duplicate:
Some three consecutive numbers sum to at least $32$ 

The integers $1, 2, \ldots, 10$ are circularly arranged in an arbitrary order. Show that there are always three successive integers in this arrangement, whose sum is at least $17.$
 A: Call the numbers, say in clockwise order, starting at some arbitrary point, $a_1$ up to $a_{10}$. Look at the sum $$(a_1+a_2+a_3)+(a_2+a_3+a_4)+ (a_3+a_4+a_5)+\cdots + (a_9+a_{10}+a_1) +(a_{10}+a_1+a_2).$$
Every number from $1$ to $10$ appears exactly $3$ times in this big sum. So the big sum is $3(1+2+\cdots+10)=165$.  If all the sums of $3$ consecutive numbers in the circular ordering were $\le 16$, the big sum could be at most $160$. But it is $165$. So at least one of the sums of $3$ consecutives in the circular ordering is $\ge 17$.
A: We will in fact prove that there is always three successive integers in this arrangement, whose sum is at least $18$.
Consider the circular arrangement of numbers starting from $1$ as follows. $$1 , a_1, a_2, \ldots, a_9$$ where $a_1,a_2,\ldots a_9 \in \{2,3,4,\ldots,10\}$.
Note that $1 + a_1 + a_2 + \cdots a_9 = 55$.
If $(a_1 + a_2 + a_3) \leq 17$ and $(a_4 + a_5 + a_6) \leq 17$ and $(a_7 + a_8 + a_9) \leq 17$, then $1 + a_1 + a_2 + \cdots a_9 \leq 52$. Which is clearly not the case, since they add up to $55$.
Hence, at least  one of $(a_1 + a_2 + a_3)$ or $(a_4 + a_5 + a_6)$ or $(a_7 + a_8 + a_9)$ is greater than $17$.
$18$ is in fact the optimal lower bound on the sum which can be seen by considering the following arrangement:
$$\{1,10,6,2,8,7,3,5,9,4\}$$ arranged circularly. The sum of any three taken in a circular fashion is at most $18$.
