A Not Quite Cyclic Inequality


A Not Quite Cyclic Inequality

Solution 1

We may set $\displaystyle \frac{a}{a+b+c}=x,\,$ $\displaystyle \frac{b}{a+b+c}=y,\,$ $\displaystyle \frac{c}{a+b+c}=z.\,$ Note that $x+y+z=1\,$ and all are non-negative. In particular, $0\le x,y,z\le 1.$ The given inequality can be rewritten as

$\displaystyle \begin{align} \frac{1}{2}\cdot\frac{(a-b)^2+(b-c)^2+(c-a)^2}{a+b+c}\le |a-b|+|b-c|&\;\Leftrightarrow\\ \frac{(x-y)^2+(y-z)^2+(z-x)^2}{2}\le |x-y|+|y-z|&\;\Leftrightarrow\\ \frac{|x-y|^2+|y-z|^2+|z-x|^2}{2}\le |x-y|+|y-z|. \end{align}$


$\displaystyle |x-y|^2\le |x-y|\le 1\\ |y-z|^2\le |y-z|\le 1\\ |z-x|^2\le |z-x|\le 1.$

It follows that

$\displaystyle \frac{|x-y|^2+|y-z|^2+|z-x|^2}{2}\le \frac{|x-y|+|y-z|+|z-x|}{2}.$

Thus, suffice it to prove that

$\displaystyle \frac{|x-y|+|y-z|+|z-x|}{2}\le |x-y|+|y-z|$

i.e., $|x-y|+|y-z|+|z-x|\le 2(|x-y|+|y-z|),\,$ or $|x-z|\le |x-y|+|y -z|,\,$ which is the triangle inequality.

Solution 2

The inequality to be proven can be written as

$2(a+b+c)(|a-b|+|b-c|) \geq (a-b)^2+(b-c)^2+(c-a)^2.$


$\displaystyle \begin{align} &p\overset{\text{def}}{=} \max\{a,b,c\},~r\overset{\text{def}}{=}\min\{a,b,c\},~q\overset{\text{def}}{=} \{a,b,c\}\setminus\{p,r\}, \\ &x\overset{\text{def}}{=} p-q,~y\overset{\text{def}}{=} q-r. \end{align}$


$\displaystyle \begin{align} &|a-b|+|b-c|\geq x+y, \\ &a+b+c=p+q+r=(q+x)+(r+y)+r\geq x+y. \end{align}$


$\displaystyle \begin{align} &2(a+b+c)(|a-b|+|b-c|) \geq 2(x+y)^2 \geq x^2+y^2+(x+y)^2 \\ &= (p-q)^2 + (q-r)^2 + (r-p)^2 = (a-b)^2 + (b-c)^2 + (c-a)^2 \end{align}$

Solution 3

A lazy proof, i.e., by brute force, using all permutations:

$\displaystyle\begin{cases} \displaystyle\frac{-a (2 b+c)+b^2+2 c^2}{a+b+c}\leq 0 & a\geq b\geq c \\ \displaystyle\frac{b^2-a c}{a+b+c}\leq 0 & a\geq c\geq b \\ \displaystyle\frac{3 \left(a^2+a b+b^2\right)}{a+b+c}+2 c\leq a+4 b & b\geq a\geq c \\ \displaystyle\frac{3 \left(a^2+a b+b^2\right)}{a+b+c}+2 c\leq a+4 b & b\geq c\geq a \\ \displaystyle\frac{b^2-a c}{a+b+c}\leq 0 & c\geq a\geq b \\ \displaystyle\frac{2 a^2-c (a+2 b)+b^2}{a+b+c}\leq 0 &c\geq b\geq a \end{cases}$


Marian Dinca has kindly messaged me the above problem, along with a solution of his. The problem is by Do Xuan Trong. Solution 2 is by Amit Itagi; Solution 3 is by M. M. Taleb


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

Copyright © 1996-2017 Alexander Bogomolny


Search by google: