# An Inequality with Permutations, II

### Problem

Let $a_1,a_2,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$ be two monotone sequences of reals - both either nondecreasing or nonincreasing. Let $\sigma\in S_n$ be a permutation of the set $\{1,2,\ldots,n\}.$ Then,

$\displaystyle\sum_{i = 1}^{n}a_ib_{n + 1 - i}\le\sum_{i = 1}^{n}a_ib_{\sigma\left(i\right)}\le\sum_{i = 1}^{n}a_ib_i.$

### Solution

Any permutation, if not a cycle itself, is a product of non-intersecting cycles: $\sigma = \sigma_{1}\sigma_{2}\ldots\sigma_{k},$ $k\ge 1.$ For simplicity, for a permutation $\tau$ of subset $I\subset\{1,2,\ldots,n\},$ let $\displaystyle\sum_{\tau}$ stand for $\displaystyle\sum_{i\in I}a_{i}b_{\tau (i)}.$ Then, clearly,

$\displaystyle\sum_{i = 1}^{n}a_ib_{\sigma (i)}=\sum_{j=1}^{k}\sum_{\sigma _{j}}.$

With this in mind, let's proceed to prove the right inequality

(*)

$\displaystyle\sum_{i = 1}^{n}a_ib_{\sigma\left(i\right)}\le\sum_{i = 1}^{n}a_ib_i.$

by induction. For $n=1,$ there is nothing to prove. For $n=2,$ the inequality becomes

$\displaystyle a_{1}b_{\sigma (1)}+a_{2}b_{\sigma (2)}\le a_{1}b_{1}+a_{1}b_{1}.$

There are two permutations of a set of two elements, the identical one for which $\sigma (1)=1$ and $\sigma (2)=2,$ the other with $\sigma (1)=2$ and $\sigma (2)=1.$ The former case reduces to double application of the basic step $(n=1).$ In the latter case,

$a_{1}b_{2}+a_{2}b_{1}\le a_{1}b_{1}+a_{2}b_{2}$

because this is equivalent to $(a_{1}-a_{2})(b_{2}-b_{1})\le 0$ which holds from the premises of the problem.

Now, let the inequality hold $n\le m.$

Consider now a permutation $\sigma$ of the set $\{1,2,\ldots,m,m+1\}$ and two monotone sequences $\{a_{i}\}$ and $\{b_{i}\},$ both either nondecreasing or nonincreasing. If $\sigma$ is the product of more than one cycle, the required inequality holds for each of the cycles by the inductive assumption and, for their product by an earlier remark. Assume then that $\sigma$ is a cycle of length $m+1.$ This in particular means that $\sigma (1)\gt 1$ and $\sigma (m+1)\gt m+1,$ implying the existence of a $j$ such that $\sigma (j)\gt j$ and also $\sigma (j)\gt\sigma (\sigma (j)).$ For simplicity, assume $\sigma (1)=3$ and $\sigma (3)=2.$ Consider then the terms $a_{1}b_{3}+a_{3}b_{2}.$ We have

$a_{1}b_{3}+a_{3}b_{2}\le a_{1}b_{2}+a_{3}b_{3}$

because the latter is equivalent to $(a_{1}-a_{3})(b_{3}-b_{2})\le 0.$ Let's define a permutation $\tau$ by $\tau (i)=\sigma (i),$ for all $i$ but $1$ and $3,$ setting $\tau (1)=2$ and $\tau (3)=3.$ (In the original terms this would mean $\tau (j)=\sigma (\sigma (j))$ and $\tau (\sigma (j))=\sigma (j).)$ So defined $\tau$ is a product of a permutation on $\{1,2,\ldots,m+1\}\setminus\{\sigma (j)\}$ and a 1-point cycle on $\{\sigma (j)\}.$ By reindexing the former we get a permutation on $\{1,2,\ldots,m\}$ for which (*) holds. Adding $a_{\sigma (j)}b_{\tau (\sigma (j))}=a_{\sigma (j)}b_{\sigma (j)}$ to both sides does not change that fact, and this proves the inductive step.

The left inequality is derived from the just proved right counterpart by replacing sequence $b_1,b_2,\ldots,b_n$ with, say, $c_{1}=-b_{n},c_{2}=-b_{n-1},\ldots,c_{n}=-b_{1},$ which is monotone in the same sence as $b_1,b_2,\ldots,b_n.$

### Acknowledgment

The problem has been posted at the Short Mathematical Idea facebook page. But the problem is not new. There is, for example, Darij Grinberg's proof from 2010.

- 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 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

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

Copyright © 1996-2017 Alexander Bogomolny

62031968 |