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1998 BMO Problems/Problem 2

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Problem

(Bogdan Enescu, Romania) If n \ge 2 is an integer and 0 < a_1 < a_2 \dotsb < a_{2n+1} are real numbers, prove the inequality: \sqrt[n]{a_1} - \sqrt[n]{a_2} + \sqrt[n]{a_3} -\dotsb - \sqrt[n]{a_{2n}} + \sqrt[n]{a_{2n+1}} < \sqrt[n]{a_1 -a_2 +a_3 - \...

Solution

Let us denote d= a_1 - a_2 + a_3 - \dotsb - a_{2n} + a_{2n+1}, and let us denote S= \sum_{i=0}^n a_{2i+1}. We note at first that d = a_1 + (a_3-a_2) + \dotsb + (a_{2n+1} - a_{2n}) > a_1 + 0 + \dotsb + 0 > 0.

Lemma. (a_1, a_3, \dotsc, a_{2n+1}) \succ (a_2, a_4, \dotsc, a_2, d)

Proof. Evidently, the two sequences have sum S. Let k be the least integer such that a_{2k} \ge d. We now note that for integers k\le j \ge n, \sum_{i=j}^n a_{2i+1} > \sum_{i=j}^n a_{2i}, and for integers 0\le j <k, \sum_{i=j}^n a_{2i+1} = S - \sum_{i=1}^{j} a_{2i-1} \ge S - \sum_{i=1}^j a_{2i}, which is the sum of the n-j smallest terms of the second sequence. Equality holds when j=0. Therefore the first sequence majorizes the second, as was to be proven. \blacksquare

Now, since the function f: x \mapsto \sqrt[n]{x} is strictly concave and our two sequences are not equal, by Karmata's Inequality \sqrt[n]{a_1} + \sqrt[n]{a_3} + \dotsb + \sqrt[n]{a_{2n+1}} < \sqrt[n]{a_2} + \dotsb + \sqrt[n]{a_{2n}} + \sqrt[n]{d}, or \sqrt[n]{a_1} - \sqrt[n]{a_2} + \sqrt[n]{a_3} - \dotsb - \sqrt[n]{a_{2n}} + \sqrt[n]{a_{2n+1}} < \sqrt[n]{d}, as desired. \blacksquare

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