Let x k be real numbers such that x k ≥ k 4 + k 2 + 1 for 1 ≤ k ≤ 2018 . Denote N = ∑ k = 1 2018 k . Consider the following inequalities. I. ∑ k = 1 2018 k x k 2 ≤ N ∑ k = 1 2018 k x k 2 II. ∑ k = 1 2018 k x k 2 ≤ N ∑ k = 1 2018 k 2 x k 2 Then,

Step-by-step solution

We have, x k ≥ k 4 + k 2 + 1 , ∀ k ∈ 1 , 2018 N ∑ k = 1 2018 k I. ∑ k = 1 2018 k x k 2 ≤ N ∑ k = 1 2018 k x k 2 We know x 1 + x 2 + x 3 + … + x n n ≤ x 1 2 + x 2 2 + x 3 2 + … + x n 2 n x 1 + x 2 + x 2 + x 3 + x 3 + x 3 + … x 2018 + x 2018    2018 times 1 + 2 + 3 + 4 + … + 2018 2 ≤ x 1 2 + x 2 2 + x 2 2 + x 3 2 + x 3 2 + x 3 2 … 1 + 2 + 3 + 4 + … + 2018 ⇒ x 1 + 2 x 2 + 3 x 3 + … + 2018 x 2018 1 + 2 + 3 +