A Cyclic Inequality in Three Variables II

Problem

A  Cyclic Inequality in Three Variables II

Solution 1

Since the inequality is homogeneous, we may assume WLOG, $a^2+b^2+c^2=1.$ First using the GM-QM inequality and then Chebyshev's inequality twice, we get

$\displaystyle\begin{align} \sum_{cycl}\frac{10a^3}{3a^2+7bc} &\ge \sum_{cycl}\left(\frac{10a^3}{3a^2+\displaystyle 7\left(\frac{b^2+c^2}{2}\right)}\right)\\ &=20\sum_{cycl}\frac{a^3}{6a^2+7b^2+7c^2}\\ &=20\sum_{cycl}\left(\frac{a^3}{7-a^2}\right)\\ &\ge\frac{20}{9}\left(\sum_{cycl}a\right)\left(\sum_{cycl}a^2\right)\left(\sum_{cycl}\frac{1}{7-a^2}\right)\\ &\ge\frac{20}{9}\left(\sum_{cycl}a\right)\left(\frac{9}{\displaystyle\sum_{cycl}(7-a^2)}\right)\\ &=\sum_{cycl}a, \end{align}$

as desired.

Solution 2

Since the inequality is homogeneous, we may assume WLOG, $abc=1.\,$ The inequality is then rewritten as $\displaystyle_{cycl}f(a)\ge 0,\,$ where

$\displaystyle f(x)\frac{10x^4}{3x^3+7}-x=\frac{7x(x^3-1)}{3x^3+7}.$

The function is convex, so that $f(x)\ge g(x)=\displaystyle\frac{21}{10}(x-1),\,$ which is its tangent at $x=1.\,$ Thus

$\displaystyle\sum_{cycl}f(a)\ge\sum_{cycl}g(a)=\frac{63}{10}-\frac{63}{10}=0,$

because $x+y+z\ge 3\sqrt[3]{xyz}=3.\,$ This is in fact the required inequality.

Solution 3

By Hölder's inequality,

$\displaystyle (a+b+c)^3 \le \sum_{cycl}\frac{a^3}{3a^2+7bc}\cdot\sum_{cycl}(3a^2+7bc)\cdot\sum_{cycl}1.$

Thus, suffice it to prove that $\displaystyle\sum_{cycl}(3a^2+7bc)\le\frac{10(a+b+c)^2}{3},\,$ or, in other words, that $a^2+b^2+c^2\ge a+b+c.\;$ But then

$\displaystyle\begin{align} \sum_{cycl}\frac{10a^3}{3a^2+7bc} &\ge \frac{10(a+b+c)^3}{\displaystyle 3\sum_{cycl}(3a^2+7bc)}\\ &\ge\frac{10(a+b+c)^3}{10(a+b+c)^2}\\ &=a+b+c. \end{align}$

Solution 4

We have a series of equivalent inequalities:

$\displaystyle\begin{align} &\frac{10a^3}{3a^2+7bc}+\frac{10b^3}{3b^2+7ca}+\frac{10c^3}{3a^2+7ab}\ge a+b+c,\\ &\sum_{cycl}\left(\frac{10a^3}{3a^2+7bc}-a\right)\ge 0,\\ &\frac{7}{2}\sum_{cycl}\frac{a(a+b)(a-c)+a(a-b)(a+c)}{3a^2+7bc}\ge 0,\\ &\frac{7}{2}\sum_{cycl}\left[\frac{a(a-b)(a+c)}{3a^2+7bc}+\frac{a(a+b)(a-c)}{3a^2+7bc}\right]\ge 0,\\ &\frac{7}{2}\sum_{cycl}\left[\frac{a(a-b)(a+c)}{3a^2+7bc}+\frac{b(b+c)(b-a)}{3b^2+7ca}\right]\ge 0,\\ &\frac{7}{2}\sum_{cycl}(a-b)\left[\frac{a(a+c)}{3a^2+7bc}-\frac{b(b+c)}{3b^2+7ca}\right]\ge 0,\\ &\frac{7}{2}\sum_{cycl}(a-b)\left[\frac{7c(a^3-b^3)+3abc(b-a)+7c^2(a^2-b^2)}{(3a^2+7bc)(3b^2+7ca)}\right]\ge 0,\\ &\frac{7}{2}\sum_{cycl}(a-b)^2\left[\frac{7c(a^2+b^2)+7c^2(a+b)+4abc}{(3a^2+7bc)(3b^2+7ca)}\right]\ge 0.\\ \end{align}$

The latter is obviously true and, so, the rest are also true.

Acknowledgment

The problem above (from the Romanian Mathematical Magazine) has been kindly communicated to me by Dan Sitaru, along with four solutions. Solution 1 is by Anas Adlany (Morroco); Solution 2 is by Imad Zak (Lebanon); Solution 3 is by Kevin Soto Palacios (Peru); Solution 3 is by Diego Alvariz (India); Solution 4 is by Soumitra Moukherjee (India).

 

Cyclic inequalities in three variables

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

Copyright © 1996-2017 Alexander Bogomolny

 62686694

Search by google: