# Dan Sitaru's Amazing, Never Ending Inequality

### Solution

Below, we shall be using repeatedly the inequality

$x^2+y^2+z^2\ge xy+yz+zx.$

For example, with $\displaystyle x=\frac{a}{b},\,$ $\displaystyle y=\frac{b}{c},\,$ $\displaystyle z=\frac{c}{a},\,$ we get

$\displaystyle \sum_{cycl}\left(\frac{a}{b}\right)^2\ge\sum_{cycl}\left(\frac{a}{b}\cdot\frac{b}{c}\right)=\sum_{cycl}\left(\frac{a}{c}\right).$

Similarly,

\displaystyle \begin{align} \sum_{cycl}\left(\frac{a}{b}\right)^4&\ge\sum_{cycl}\left(\frac{a}{b}\cdot\frac{b}{c}\right)^2\\ &=\sum_{cycl}\left(\frac{a}{c}\right)^2\ge\sum_{cycl}\left(\frac{a}{c}\cdot\frac{b}{a}\right)\\ &=\sum_{cycl}\left(\frac{b}{c}\right),\\ \sum_{cycl}\left(\frac{a}{b}\right)^8&\ge\sum_{cycl}\left(\frac{a}{b}\cdot\frac{b}{c}\right)^4\\ &=\sum_{cycl}\left(\frac{a}{c}\right)^4\ge\sum_{cycl}\left(\frac{a}{c}\cdot\frac{b}{a}\right)^2\\ &=\sum_{cycl}\left(\frac{b}{c}\right)^2\ge\sum_{cycl}\left(\frac{b}{c}\cdot\frac{c}{a}\right)\\ &=\sum_{cycl}\left(\frac{c}{a}\right). \end{align}

We now only need to multiply the three inequalities.

### Extra

The problem admits multiple variations. For example, from

\displaystyle \begin{align} \sum_{cycl}\left(\frac{a}{b}\right)^4&\ge\sum_{cycl}\left(\frac{a}{b}\cdot\frac{b}{c}\right)^2\\ &=\sum_{cycl}\left(\frac{a}{c}\right)^2\ge\sum_{cycl}\left(\frac{a}{c}\cdot\frac{c}{b}\right)\\ &=\sum_{cycl}\left(\frac{a}{b}\right), \end{align}

we get that, for $k\ge 2,\,$ $\displaystyle \sum_{cycl}\left(\frac{a}{b}\right)^{2^k}\ge\sum_{cycl}\left(\frac{a}{b}\right)\,$ and thus

$\displaystyle \prod_{k=2}^n\left[\sum_{cycl}\left(\frac{a}{b}\right)^{2^k}\right]\ge\left[\sum_{cycl}\left(\frac{a}{b}\right)\right]^{n-1}.$

By the same token, the original inequality could have been written as

$\displaystyle \small{\sum_{cycl}\left(\frac{a}{b}\right)^2\cdot\sum_{cycl}\left(\frac{a}{b}\right)^4\cdot\sum_{cycl}\left(\frac{a}{b}\right)^8\ge\sum_{cycl}\left(\frac{a}{c}\right)\cdot\sum_{cycl}\left(\frac{c}{b}\right)\cdot\sum_{cycl}\left(\frac{b}{a}\right)}$

or as

$\displaystyle \sum_{cycl}\left(\frac{a}{b}\right)^2\cdot\sum_{cycl}\left(\frac{a}{b}\right)^4\cdot\sum_{cycl}\left(\frac{a}{b}\right)^8\ge\left[\sum_{cycl}\left(\frac{a}{c}\right)\right]^3$

and in general

$\displaystyle \prod_{k=1}^n\left[\sum_{cycl}\left(\frac{a}{b}\right)^{2^k}\right]\ge\left[\sum_{cycl}\left(\frac{a}{c}\right)\right]^{n}.$

### Acknowledgment

Dan Sitaru has kindly posted the above problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page and later emailed me his solution of this amazing problem in a LaTeX file.

Copyright © 1996-2018 Alexander Bogomolny

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