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The numbers $1, 2, \dots , 10$ are arranged in a circle in some order. Show that it is always possible to find three adjacent numbers whose sum is at least $17$, irrespective of the ordering.
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${1,2,3,4,5,6,7,8,9,10}$

sum of first 10 natural numbers = $\frac{10 . 11}{2}$ = 55

we need to show that there exist some combination of 3 numbers so that sum of those 3 numbers = 17

let us assume {1,a2,a3,a4,a5,a6,a7,a8,a9,a10

sum of a2 + a3+........+ a9 = 55-1 = 54

if we choose any 3 numbers from a2 to a9 and try to evenly distribute the sum 54 between them,

$\frac{54}{3} = 18$ 

this shows that how evenly we pick numbers from 2 to 10 it can never go below 18 for some combination of a2 to a9.

18 is the optimal lower bound.

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