A Cyclic Inequality in Three Variables XXV


A   Cyclic  Inequality in   Three Variables  XXV

Solution 1

By the AM-GM inequality, $a-\sqrt{ab}+b\ge \sqrt{ab}\,$ and $a^2-ab+b^2\ge ab.\,$ It thus follows that

$\displaystyle \sum_{cycl} (a-\sqrt{ab}+b)^2\cdot\sum_{cycl}(a^2-ab+b^2)^2\ge\sum_{cycl}ab\cdot\sum_{cycl}a^2b^2.$

Again, with the AM-GM inequality, $\displaystyle \sum_{cycl}ab\ge 3\sqrt[3]{a^2b^2c^2}\,$ and $\displaystyle \sum_{cycl}a^2b^2\ge 3\sqrt[3]{a^4b^4c^4}\,$ so that

$\displaystyle \sum_{cycl}ab\cdot\sum_{cycl}a^2b^2\ge 9\sqrt[3]{a^2b^2c^2}\cdot\sqrt[3]{a^4b^4c^4}=9a^2b^2c^2.$

Solution 2

With Bergstrom's inequality,

$\displaystyle\begin{align} \sum_{cycl} (a-\sqrt{ab}+b)^2&\ge\frac{\displaystyle \left(2\sum_{cycl}a-\sqrt{ab}\right)^2}{3}\\ &\ge \frac{\displaystyle \left(\sum_{cycl}a\right)^2}{3}\\ &\ge \frac{\displaystyle 3(\sum_{cycl}ab)}{3}\ge \sum_{cycl}ab \end{align}$

and, similarly, $\displaystyle \sum_{cycl} (a^2-ab+b^2)^2\ge \sum_{cycl}a^2b^2.\,$ Further,

$\displaystyle \begin{align} \sum_{cycl}ab\cdot\sum_{cycl}a^2b^2&\ge 3\sum_{cycl}a^3b^2c\\ &\ge 9\sqrt[3]{a^6b^6c^6}=9a^2b^2c^2. \end{align}$


Dan Sitaru has kindly posted the problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page and later communicated the above proof in a LaTeX file. Solution 1 is by Kevin Soto Palacios and, independently, by Seyran Ibrahimov; Solution 2 is by Sanong Hauerai and a similar solution by has been posted by Abdur Rahman.


Cyclic inequalities in three variables

|Contact| |Up| |Front page| |Contents| |Algebra|

Copyright © 1996-2017 Alexander Bogomolny


Search by google: