# Cyclic Inequality in Four Variables

### Problem

### Solution

First, by the AM-GM inequality, a assuming, WLOG, that $abcd=1$ we get:

(1)

$\displaystyle \sum_{cycl}a^2\ge 4\sqrt[4]{(abcd)^2}=4$

and,

(1')

$\displaystyle \sum_{all}ab\ge 6\sqrt[6]{(abcd)^3}=6.$

Now, by Bergstrom's inequality,

(2)

$\displaystyle \sum_{cycl}\frac{a^3}{bcd}=\sum_{cycl}\frac{a^4}{abcd}\ge\frac{\displaystyle \left(\sum_{cycl}a^2\right)^2}{4abcd}=\frac{1}{4}\left(\sum_{cycl}a^2\right)^2.$

Further, because of (1'),

(3)

$\displaystyle 1+\frac{3\displaystyle \sum_{cycl}a^2}{\displaystyle \sum_{all}ab}\le 1+\frac{1}{2}\sum_{cycl}a^2.$

Thus, suffice it to prove that

(4)

$\displaystyle \frac{3}{4}\cdot\frac{1}{4}\left(\sum_{cycl}a^2\right)^2\ge 1+\frac{1}{2}\sum_{cycl}a^2.$

With $\displaystyle x=\sum_{cycl}a^2,$ this is equivalent to $3x^2-8x-16\ge 0.$ This is true for $x\ge 4$ which holds due to (1).

Equality is attained for $a=b=c=d.$

### Acknowledgment

- This is a problem by Sladjan Stankovik posted at the mathematical inequalities facebook group. Additional solutions can be found at the link.

[an error occurred while processing this directive]

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

Copyright © 1996-2018 Alexander Bogomolny