# Two Cyclic Inequalities

### Introduction

Two problems discussed below have a curious story to tell. The first was proposed by the Moscow mathematician D. P. Mavlo and published in *Matematica*, No. 4, 1987. It was intended to be a difficult problem whose solution relied on calculus. But a short elementary solution has been submitted by a Bulgarian boy Zhivko Georgiev, a ninth-grader at the time. Zhivko's solution is given below.

The second problem has been published next year in journal *Kvant* (No. 11-12, 1988, problem M1136). It was dedicated to the centennial of the American Mathematical Society and was not supposed to be easy, but still elementary. The solution to the problem has been published in Kvant, No. 5, 1989. It was observed (Savchev and Andreescu, see the references at the bottom of the page) that the concise solution sounded more like a hint. But another solution can be obtained by simple substitution from the first problem. Thus it must be possible to proceed backwards and obtain another elementary solution to the latter. (Well, not that I did not try...)

### Problem 1

Let $a,b,c$ be positive numbers. Prove that

$\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\ge\frac{3}{1+abc},$

and the equality occurs if and only if $a=b=c=1.$

### Solution 1 to Problem 1

Observe that

$\begin{align}\displaystyle \frac{1}{a(1+b)}+\frac{1}{1+abc}&=\frac{1+a+ab+abc}{a(1+b)(1+abc)}\\ &=\frac{1}{1+abc}\bigg[\frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}\bigg]. \end{align}$

And similarly

$\begin{align}\displaystyle \frac{1}{b(1+c)}+\frac{1}{1+abc}&=\frac{1+b+bc+abc}{b(1+c)(1+abc)}\\ &=\frac{1}{1+abc}\bigg[\frac{1+b}{b(1+c)}+\frac{c(1+a)}{1+c}\bigg], \end{align}$

$\begin{align}\displaystyle \frac{1}{c(1+a)}+\frac{1}{1+abc}&=\frac{1+c+ac+abc}{c(1+a)(1+abc)}\\ &=\frac{1}{1+abc}\bigg[\frac{1+c}{c(1+a)}+\frac{a(1+b)}{1+a}\bigg]. \end{align}$

Before adding all three, introduce

$\displaystyle x=\frac{1+a}{a(1+b)},\space y=\frac{1+b}{b(1+c)},\space z=\frac{1+c}{c(1+a)}.$

With these, after adding up, what we get in the left-hand side

$\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}+\frac{3}{1+abc} $

whereas on the right we get

$\displaystyle \frac{1}{1+abc}\left[x+\frac{1}{x}+y+\frac{1}{y}+z+\frac{1}{z}\right]\ge \frac{6}{1+abc}, $

because for any positive $u,$ $\displaystyle u+\frac{1}{u}\ge 2,$ with equality only when $u=1.$

Putting everything together,

$\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}+\frac{3}{1+abc} \ge \frac{6}{1+abc}, $

i.e., the inequality we set out to prove.

### Problem 2

Prove the inequality

$\begin{align} \displaystyle 3+(A+M+S)+\bigg(\frac{1}{A}+\frac{1}{M}+\frac{1}{S}\bigg)&+\bigg(\frac{A}{M}+\frac{M}{S}+\frac{S}{A}\bigg)\\ &\ge\frac{3(A+1)(M+1)(S+1)}{AMS+1} \end{align}$

for all positive $A,M,S.$

### Solution 1 to Problem 2

First, trivially, $3+(A+M+S)=(1+A)+(1+M)+(1+S).$ Next, observe that

$\displaystyle (1+A)+\frac{1}{M}+\frac{A}{M}=\frac{(1+A)(1+M)}{M},$ etc.

So, the left-hand side transforms into

$\displaystyle\frac{(1+A)(1+M)}{M}+\frac{(1+M)(1+S)}{S}+\frac{(1+S)(1+A)}{A},$

such that after division by $(1+A)(1+M)(1+S)$ the inequality becomes

$\displaystyle\frac{1}{M(1+S)}+\frac{1}{S(1+A)}+\frac{1}{A(1+M)}\ge\frac{3}{1+AMS},$

which is the earlier inequality.

### Solution 2 to Problem 2

Here's the solution published in *Kvant*:

The proof reduces to a rather artificial (but also ingenious!) algebraic manipulation. Multiply the difference of the left-hand and the right-hand sides by $AMS(1+AMS),$ regroup the summands, and you will obtain

$AM(M+1)(SA-1)^{2}+MS(S+1)(AM-1)^{2}+SA(A+1)(MS-1)^{2}$

a non-negative quantity. This proof was found by P. H. Diananda from the University of Singapore.

### References

- S. Savchev, T. Andreescu,
*Mathematical Miniatures*, MAA, 2003, pp. 101-103

- An Inequality for Grade 8
- An Extension of the AM-GM Inequality
- Schur's Inequality
- Newton's and Maclaurin's Inequalities
- Rearrangement Inequality
- Chebyshev Inequality
- Jensen's Inequality
- Muirhead's Inequality
- Bergström's inequality
- Radon's Inequality and Applications
- Jordan and Kober Inequalities, PWW
- A Mathematical Rabbit out of an Algebraic Hat
- An Inequality With an Infinite Series
- An Inequality: 1/2 * 3/4 * 5/6 * ... * 99/100 less than 1/10
- A Low Bound for 1/2 * 3/4 * 5/6 * ... * (2n-1)/2n
- An Inequality: Easier to prove a subtler inequality
- Inequality with Logarithms
- An inequality: 1 + 1/4 + 1/9 + ... less than 2
- Inequality with Harmonic Differences
- An Inequality by Uncommon Induction
- Hlawka's Inequality
- An Inequality in Determinants
- Application of Cauchy-Schwarz Inequality
- An Inequality from Tibet
- An Inequality with Constraint
- An Inequality from Morocco
- An Inequality for Mixed Means
- An Inequality in Integers
- An Inequality in Integers II
- An Inequality in Integers III
- An Inequality with Exponents
- Exponential Inequalities for Means
- A Simple Inequality in Three Variables
- An Asymmetric Inequality
- Linear Algebra Tools for Proving Inequalities
- An Inequality with a Generic Proof
- A Generalization of an Inequality from a Romanian Olympiad
- Area Inequality in Trapezoid
- Improving an Inequality
- RomanoNorwegian Inequality
- Inequality with Nested Radicals II
- Inequality with Powers And Radicals
- Inequality with Two Minima
- Simple Inequality with Many Faces And Variables
- An Inequality with Determinants
- An Inequality with Determinants II
- An Inequality with Determinants III
- An Inequality with Determinants IV
- An Inequality with Determinants V
- An Inequality with Determinants VI
- An Inequality with Determinants VII
- An Inequality in Reciprocals
- An Inequality in Reciprocals II
- An Inequality in Reciprocals III
- Monthly Problem 11199
- A Problem from the Danubius Contest 2016
- A Problem from the Danubius-XI Contest
- An Inequality with Integrals and Rearrangement
- An Inequality with Cot, Cos, and Sin
- A Trigonometric Inequality from the RMM
- An Inequality with Finite Sums
- Hung Viet's Inequality
- Hung Viet's Inequality II
- Hung Viet's Inequality III
- Inequality by Calculus
- Dorin Marghidanu's Calculus Lemma
- An Area Inequality
- A 4-variable Inequality from the RMM
- An Inequality from RMM with Powers of 2
- A Cycling Inequality with Integrals
- A Cycling Inequality with Integrals II
- An Inequality with Absolute Values
- An Inequality from RMM with a Generic 5
- An Elementary Inequality by Non-elementary Means
- Inequality in Quadrilateral
- Marian Dinca's Refinement of Nesbitt's Inequality
- An Inequality in Cyclic Quadrilateral
- An Inequality in Cyclic Quadrilateral II
- An Inequality in Cyclic Quadrilateral III
- An Inequality in Cyclic Quadrilateral IV
- Inequality with Three Linear Constraints
- Inequality with Three Numbers, Not All Zero
- An Easy Inequality with Three Integrals
- Divide And Conquer in Cyclic Sums
- Wu's Inequality
- A Cyclic Inequality in Three Variables
- Dorin Marghidanu's Inequality in Complex Plane
- Dorin Marghidanu's Inequality in Integer Variables
- Dorin Marghidanu's Inequality in Many Variables
- Dorin Marghidanu's Inequality in Many Variables Plus Two More
- Dorin Marghidanu's Inequality with Radicals
- Dorin Marghidanu's Light Elegance in Four Variables
- Dorin Marghidanu's Spanish Problem
- Two-Sided Inequality - One Provenance
- An Inequality with Factorial
- Wonderful Inequality on Unit Circle
- Quadratic Function for Solving Inequalities
- An Inequality Where One Term Is More Equal Than Others
- An Inequality and Its Modifications
- Complicated Constraint - Simple Inequality
- Distance Inequality
- Two Products: Constraint and Inequality
- The power of substitution II: proving an inequality with three variables
- Algebraic-Geometric Inequality
- One Inequality - Two Domains
- Radicals, Radicals, And More Radicals in an Inequality
- An Inequality in Triangle and In General
- Cyclic Inequality with Square Roots
- Dan Sitaru's Cyclic Inequality In Many Variables
- An Inequality on Circumscribed Quadrilateral
- An Inequality with Fractions
- An Inequality with Complex Numbers of Unit Length
- An Inequality with Complex Numbers of Unit Length II
- Le Khanh Sy's Problem
- An Inequality Not in Triangle
- An Acyclic Inequality in Three Variables
- An Inequality with Areas, Norms, and Complex Numbers
- Darij Grinberg's Inequality In Three Variables
- Small Change Makes Big Difference
- Inequality with Two Variables? Think Again
- A Problem From a Mongolian Olympiad for Grade 11
- Sitaru--Schweitzer Inequality
- An Inequality with Cyclic Sums And Products
- Problem 1 From the 2016 Pan-African Math Olympiad
- An Inequality with Integrals and Radicals
- Twin Inequalities in Four Variables: Twin 1
- Twin Inequalities in Four Variables: Twin 2
- Simple Inequality with a Variety of Solutions
- A Partly Cyclic Inequality in Four Variables
- Dan Sitaru's Inequality by Induction
- An Inequality in Three (Or Is It Two) Variables
- An Inequality in Four Weighted Variables
- An Inequality in Fractions with Absolute Values
- Inequalities with Double And Triple Integrals
- An Old Inequality
- Dan Sitaru's Amazing, Never Ending Inequality
- Leo Giugiuc's Exercise
- Another Inequality with Logarithms, But Not Really
- A Cyclic Inequality of Degree Four
- An Inequality Solved by Changing Appearances
- Distances to Three Points on a Circle
- An Inequality with Powers And Logarithm
- Four Integrals in One Inequality
- Same Integral, Three Intervals
- Dorin Marghidanu's Inequality with Generalization
- Dan Sitaru's Inequality with Three Related Integrals and Derivatives
- An Inequality in Two Or More Variables
- An Inequality in Two Or More Variables II
- A Not Quite Cyclic Inequality
- Dan Sitaru's Inequality: From Three Variables to Many in Two Ways
- An Inequality with Sines But Not in a Triangle
- An Inequality with Angles and Integers
- Sladjan Stankovik's Inequality In Four Variables
- An Inequality with Two Pairs of Triplets
- A Refinement of Turkevich's Inequality
- Dan Sitaru's Exercise with Pi and Ln
- Problem 4165 from Crux Mathematicorum
- Leo Giugiuc's Cyclic Quickie in Four Variables

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

Copyright © 1996-2017 Alexander Bogomolny

62336826 |