# Dorin Marghidanu's Example II

### Problem

### Solution 1

With relation $\displaystyle s=\sum_{k=1}^na_k=\sum_{k=1}^na_{\sigma (k)},\,$ $\sigma\in S_n,\,$ and by Bergström's Inequality, we have

$\displaystyle\sum_{k=1}^n\frac{a_k^2}{a_{\sigma (k)}-1} \ge \frac{\displaystyle \left(\sum_{k=1}^na_k\right)^2}{\displaystyle \sum_{k=1}^n(a_{\sigma (k)}-1)}=\frac{s^2}{s-n}.$

It remains to be shown that $\displaystyle \frac{s^2}{s-n}\ge 4n\,$ which is equivalent to $(s-2n)^2\ge 0.\,$ Equality occurs iff $s=2n,\,$ which, together with conditions of equality in Bergstrom's inequality, leads to

$\displaystyle\frac{a_1}{a_{\sigma (1)}-1}=\frac{a_2}{a_{\sigma (2)}-1}=\ldots=\frac{a_n}{a_{\sigma (n)}-1}=\frac{s}{s-n}=\frac{2n}{n}=2,$

so that $a_1=a_2=\ldots=a_n=2.$

With $n=5\,$ and an identity permutation the problem appeared at the RMO (Regional Mathematics Olympiad), 2013.

### Solution 2

By the Rearrangement inequality, suffices it to prove the result for the identity permutation, i.e., $\sigma(k) = k,\,$ for every $k.\,$ Next (assuming, as stated, that $\sigma =e)\,$ prove that each term of the sum is greater that or equal to $4.\,$ In fact, $0 \le (x-2)^2 = x^2 - 4x + 4,\,$ which is equivalent to $\displaystyle \frac{x^2}{x-1} \ge 4,\,$ and the result follows.

### Acknowledgment

Dorin Marghidanu has kindly posted the elegant problem above, with solution (Solution 1) that utilized Bergström's inequality, at the CutTheKnotMath facebook page. He later added another solution (Solution 2), this by Miguel A. Lerma.

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