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Algebra Level 5

Consider k k positive real numbers: a 1 , a 2 , a 3 , , a k a_1,a_2,a_3,\ldots , a_{k} such that i = 1 k a i = 2016 \displaystyle\prod_{i=1}^{k} a_i=2016 and i = 1 k a i k = 2016 k \displaystyle\sum_{i=1}^{k} a_i^{k}=2016 k for some positive integer k k .

Then, find the value of i = 1 2014 a i a i + 1 + a i + 2 \displaystyle\sum_{i=1}^{2014} \dfrac{a_i}{a_{i+1}+a_{i+2}} .


The answer is 1007.

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1 solution

Nihar Mahajan
Apr 25, 2016

By doing A M G M AM-GM inequality, we can easily establish i = 1 k a i k 2016 k \displaystyle\sum_{i=1}^{k} a_i^{k}\geq 2016k but its is given that it exactly equals 2016 k 2016k , thus we consider the equality case giving a 1 = a 2 = a 3 = = a k a_1=a_2=a_3=\dots=a_{k} . Thus our required sum i = 1 2014 a i a i + 1 + a i + 2 = i = 1 2014 a i 2 a i = 1 2 i = 1 2014 1 = 2014 2 = 1007 \displaystyle\sum_{i=1}^{2014} \dfrac{a_i}{a_{i+1}+a_{i+2}}= \displaystyle\sum_{i=1}^{2014} \dfrac{a_i}{2a_i} = \dfrac{1}{2}\displaystyle\sum_{i=1}^{2014} 1 = \dfrac{2014}{2}=\boxed{1007}

Moderator note:

Good usage of the equality case of AM-GM.

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