# Simple Nameless Inequality

### Solution 1

We'll use Bergstrom's inequality:

\displaystyle\begin{align}\sum_{k=1}^n\frac{S}{S_k}&=S\sum_{k=1}\frac{1}{S_k}\\ &\ge S\cdot\frac{(1+1+\ldots+1)^2}{nS-S}=S\cdot\frac{n^2}{(n-1)S}\\ &=\frac{n^2}{n-1}. \end{align}

The equality is attained iff all $S_k,$ i.e., all $x_k$ are equal.

### Solution 2

By the Cauchy-Schwarz inequality,

\displaystyle\begin{align}n^2&\le\left(\sum_{k=1}^n\frac{S}{S_k}\right)\left(\sum_{k=1}^n\frac{S_k}{S}\right)=\left(\sum_{k=1}^n\frac{S}{S_k}\right)\left(\sum_{k=1}^n\left(1-\frac{x_k}{S}\right)\right)\\ &=\left(\sum_{k=1}^n\frac{S}{S_k}\right)\left(n-\frac{\displaystyle \sum_{k=1}^ax_k}{S}\right) =\left(\sum_{k=1}^n\frac{S}{S_k}\right)n-1), \end{align}

with equality iff $\displaystyle \sqrt{\frac{S}{S_k}}$ is constant multiple of $\displaystyle \sqrt{\frac{S_k}{S}},$ i.e., itself constant.

### Solution 3

Let's invoke the AM-AM inequality: $\displaystyle \frac{n}{\displaystyle \sum_{k=1}^n\frac{1}{a_k}}\le\frac{1}{n}\sum_{k=1}^na_k.$ Set $\displaystyle a_k=\frac{S}{S_k},$ so that $\displaystyle \sum_{k=1}^na_k=n-1.$ Then the result is immediate.

### Acknowledgment

This is problem 146 from Five Hundred Mathematical Challenges by E. J. Barbeau et all (MAA, 1995). Solutions 2 and 3 come from the book.

Copyright © 1996-2018 Alexander Bogomolny

 63580130

Search by google: