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Algebra Level 4

Find the minimum value of ∑ k = 1 2014 a k 2014 + ∑ k = 1 2014 1 a k 2014 \displaystyle\sum_{k=1}^{2014} a_k^{2014} + \sum_{k=1}^{2014} \frac1{a_k^{2014}} where a 1 , a 2 , ⋯ , a 2014 > 0 a_1, a_2, \cdots, a_{2014} > 0 .

Inspired by Calvin's comment on one of the solutions of this post


The answer is 4028.

Kenny Lau by Kenny Lau , 22, Hong Kong

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1 solution

Sujoy Roy
Dec 2, 2014

We know, a k + 1 a k ≥ 0 a_{k}+\frac{1}{a_{k}} \ge 0 for a i > 0 a_{i}>0 .

So, ∑ k = 1 2014 ( a k + 1 a k ) ≥ 2 ∗ 2014 = 4028 \sum_{k=1}^{2014}(a_{k}+\frac{1}{a_{k}}) \ge 2*2014=\boxed{4028}

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