Consider k positive real numbers: a 1 , a 2 , a 3 , … , a k such that i = 1 ∏ k a i = 2 0 1 6 and i = 1 ∑ k a i k = 2 0 1 6 k for some positive integer k .
Then, find the value of i = 1 ∑ 2 0 1 4 a i + 1 + a i + 2 a i .
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Good usage of the equality case of AM-GM.
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By doing A M − G M inequality, we can easily establish i = 1 ∑ k a i k ≥ 2 0 1 6 k but its is given that it exactly equals 2 0 1 6 k , thus we consider the equality case giving a 1 = a 2 = a 3 = ⋯ = a k . Thus our required sum i = 1 ∑ 2 0 1 4 a i + 1 + a i + 2 a i = i = 1 ∑ 2 0 1 4 2 a i a i = 2 1 i = 1 ∑ 2 0 1 4 1 = 2 2 0 1 4 = 1 0 0 7