# 1998 BMO Problems/Problem 2

## 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: $$\sqrt[n]{a_1} - \sqrt[n]{a_2} + \sqrt[n]{a_3} -\dotsb - \sqrt[n]{a_{2n}} + \sqrt[n]{a_{2n+1}} < \sqrt[n]{a_1 -a_2 +a_3 - \dotsb -a_{2n} + a_{2n+1}} .$$

## 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 , $$\sum_{i=j}^n a_{2i+1} = S - \sum_{i=1}^{j} a_{2i-1} \ge S - \sum_{i=1}^j a_{2i},$$ 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. $\blacksquare$