An Easy Cyclic Inequality And a Remark
Problem
Solution 1
For one,
$\displaystyle \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\ge \frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1.$
Also note that $\displaystyle \frac{x}{y}\le\frac{x+u}{y+u},$ $x\le y,$ $u\ge 0.$ Using that
$\displaystyle \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\le \frac{a+c}{a+b+c}+\frac{b+a}{a+b+c}+\frac{c+b}{a+b+c}=2.$
Rather obviously both inequalities are strict, even allowing for one of the variables be zero.
Equivalences
It is remarkable to see that the left side inequality is equivalent to the right side inequality. Indeed, for the left inequality
$\begin{align}\displaystyle &1\le\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}&\Leftrightarrow\\ &1-\frac{a}{a+b}\le\frac{b}{b+c}+\frac{c}{c+a}&\Leftrightarrow\\ &\frac{b}{a+b}\le\frac{b}{b+c}+\frac{c}{c+a}. \end{align}$
For the right side, we have
$\begin{align}\displaystyle &\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\le 2&\Leftrightarrow\\ &\frac{a}{a+b}+\left(1-\frac{c}{b+c}\right)+\left(1-\frac{a}{c+a}\right)\le 2&\Leftrightarrow\\ &\frac{a}{a+b}\le\frac{c}{b+c}+\frac{a}{c+a}. \end{align}$
In the published remark, probably by the Problems Editors, it is claimed that the last two inequalities are also equivalent:
$\begin{align}\displaystyle &\frac{b}{a+b}\le\frac{b}{b+c}+\frac{c}{c+a}&\Leftrightarrow\\ &\frac{a}{a+b}\le\frac{c}{b+c}+\frac{a}{c+a}. \end{align}$
"The above inequalities differ only by a cyclic shift." This is not true. Each of the inequalities are invariant under a cyclic shift (this is why they are cyclic) but differ from each other. We have
$\begin{align}\displaystyle &\frac{b}{a+b}\le\frac{b}{b+c}+\frac{c}{c+a}&\Leftrightarrow\\ &(a+b)[b(c+a)+c(b+c)]-b(b+c)(c+a)\ge 0&\Leftrightarrow\\ &a^2b+b^2c+c^2a+abc\ge 0.\\ &\frac{a}{a+b}\le\frac{c}{b+c}+\frac{a}{c+a}.&\Leftrightarrow\\ &(a+b)[c(c+a)+a(b+c)]-a(b+c)(c+a)\ge 0&\Leftrightarrow\\ &a^2c+c^2b+b^2a+abc\ge 0. \end{align}$
The expressions have different values, e.g., for $a=1,$ $b=2,$ $c=3.$
Solution 2
Let $x=b/a$, $y=c/b$, $z=a/c$. This transforms the inequality to
$\displaystyle \begin{align} &2\geq f(x,y,z)\geq 1, \\ &f(x,y,z)=\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}; \end{align}$
under the constraint $xyz=1$.
$\displaystyle \begin{align} f(x,y,z)&=\frac{(1+y)(1+z)+(1+z)(1+x)+(1+x)(1+y)}{(1+x)(1+y)(1+z)} \\ &=\frac{3+2(x+y+z)+(xy+yz+zx)}{1+(x+y+z)+(xy+yz+zx)+xyz} \\ &=\frac{3+2(x+y+z)+(xy+yz+zx)}{2+(x+y+z)+(xy+yz+zx)} \\ &=1+\left[\frac{1+(x+y+z)}{2+(x+y+z)+(xy+yz+zx)}\right] \\ &=2-\left[\frac{1+(xy+yz+zx)}{2+(x+y+z)+(xy+yz+zx)}\right]. \end{align}$
The box brackets on the last two lines are non-negative. Thus,
$2\geq f(x,y,z) \geq 1.$
Solution 3
The inequality being homogenous in $a$, $b$, and $c$, WLOG, let $a+b+c=1$. Thus,
$\displaystyle \begin{align} \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} &=\frac{a(a+b+c)}{a+b}+\frac{b(a+b+c)}{b+c}+\frac{c(a+b+c)}{c+a} \\ &=(a+b+c)+\left[\frac{ac}{a+b}+\frac{ba}{b+c}+\frac{cb}{c+a}\right] \\ &=1+\left[\frac{ac}{a+b}+\frac{ba}{b+c}+\frac{cb}{c+a}\right] \geq 1. \end{align}$
By a similar argument, or by interchanging the dummy variables $b$ and $c$,
$\displaystyle \frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a} \geq 1.$
Both terms are $\geq 1$ and the terms add up to $3$. Thus, both terms are $\leq 2$.
Acknowledgment
This is Problem 4196 from the Canadian Crux Mathematicorum (Vol. 43(10), December 2017); the problem was posed by Leo Giugiuc and Dan Sitaru. The latter has kindly brought the problem to my attention. Solution 1 is by Adnan Ali; Solutions 2 and 3 are by Amit Itagi.
Cyclic inequalities in three variables
- ab + bc + ca does not exceed aa + bb + cc
- A Cyclic Inequality in Three Variables $\left(\displaystyle\frac{a^3}{b^2(5a+2b)}+\frac{b^3}{c^2(5b+2c)}+\frac{c^3}{a^2(5c+2a)}\ge\frac{3}{7}\right)$
- A Cyclic Inequality in Three Variables II $\left(\displaystyle\frac{10a^3}{3a^2+7bc}+\frac{10b^3}{3b^2+7ca}+\frac{10c^3}{3a^2+7ab}\ge a+b+c\right)$
- A Cyclic Inequality in Three Variables III $\left(\displaystyle\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\ge 2\right)$
- A Cyclic Inequality in Three Variables IV $\left(\displaystyle 2\sum_{cycl}(a+b)^3+5\sum_{cycl}a^3\ge 21\sum_{cycl}a^2b\right)$
- A Cyclic Inequality in Three Variables V $\left(\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{\sqrt{3(a^2+b^2+c^2)}\cdot (a+b+c)}{ab+bc+ca}\right)$
- A Cyclic Inequality in Three Variables VI $\left(\displaystyle \frac{2(a+b+c)}{abc}\ge\sum_{cycl}\left(\sqrt{\frac{a+b}{2ac}}+\sqrt{\frac{2a}{c(a+b)}}\right)\right)$
- A Cyclic Inequality in Three Variables VII $\left(\displaystyle\sum_{cycl}x\sqrt{x^2z^2+y^4}\ge\sqrt{2}\sum_{cycl}xz\sqrt{yz}\right)$
- A Cyclic Inequality in Three Variables VIII $\left(\displaystyle\sum_{cycl}(x^2+y^2)z+\sum_{cycl}\frac{xy}{(x+y)^2}\ge 27xyz\right)$
- A Cyclic Inequality in Three Variables IX $\left(\displaystyle 9\left(\sum_{cycl}\frac{x^2}{y^2}\right)^2\ge 8\left(\sum_{cycl}\frac{x}{y}\right)\left(\sum_{cycl}\frac{x^3}{y^3}-3\right)\right)$
- A Cyclic Inequality in Three Variables X $\left(\displaystyle\sum_{cycl}\frac{1}{(a+1)^3}+4\sum_{cycl}\frac{1}{(a+1)^4}\ge\frac{9}{8}\right)$
- A Cyclic Inequality in Three Variables XI $\left(\displaystyle\sum_{cycl}\frac{1}{(a^2-ab+b^2)(b^2-bc+c^2)}\le\sum_{cycl}\frac{1}{a^4}\right)$
- A Cyclic Inequality in Three Variables XII $\left(\displaystyle\left(\sum_{cycl}\frac{1}{(a^2-ab+b^2)^6}\right)^2\le 3\sum_{cycl}\left(\frac{a+b}{a^2+b^2}\right)^{24}\right)$
- A Cyclic Inequality in Three Variables XIII $\left(\displaystyle\sum_{cycl}\frac{a^2+b^2}{a+b}+11\sum_{cycl}\frac{ab}{a+b}\gt 6\sum_{cycl}\sqrt{ab}\right)$
- A Cyclic Inequality in Three Variables XIV $\left(\displaystyle\sum_{cycl}\frac{xy}{xy+y^2+zx}\le 1\right)$
- A Cyclic Inequality in Three Variables XV $\left(\displaystyle \frac{a(a^2+b^2)}{a^3+b^3}+\frac{b(b^2+c^2)}{b^3+c^3}+\frac{c(c^2+a^2)}{c^3+a^3}\leq \sqrt{\frac{a}{b}}+\sqrt{\frac{b}{c}}+\sqrt{\frac{c}{a}}\right)$
- A Cyclic Inequality in Three Variables XVI $\left(\displaystyle \sum_{cycl}|(a+b)(1-ab)|\lt\frac{3}{2}+\sum_{cycl}a^2+\frac{1}{2}\sum_{cycl}a^4\right)$
- A Cyclic Inequality in Three Variables XVII $\left(\displaystyle \left(\sum_{cycl}\frac{x^2}{y^2}\right)^5 \ge 9\left(\sum_{cycl}\frac{x^3}{y^2z}\right)\left(\sum_{cycl}\frac{x}{\sqrt{yz}}\right)\left(\sum_{cycl}\frac{y}{z}\right)\right)$
- A Cyclic Inequality in Three Variables XVIII $\left(\displaystyle \left(\sum_{cycl}\sqrt{ab}\right)^6 \le 27\prod_{cycl}(a^2+ab+b^2)\right)$
- A Cyclic Inequality in Three Variables XIX $\left(\displaystyle \frac{x}{(y+z)^3}+\frac{y}{(z+x)^3}+\frac{z}{(x+y)^3}\ge\frac{27}{8(x+y+z)^2}\right)$
- A Cyclic Inequality in Three Variables XX $\left(\displaystyle 5\sum_{cycl}\sqrt{ab}\le\sum_{cycl}\sqrt[4]{(a+4b)(2a+3b)(3b+2a)(4a+b)}\le 5\right)$
- A Cyclic Inequality in Three Variables XXI $\left(\displaystyle \frac{abc}{7\sqrt{7}}\le\prod_{cycl}\frac{a^2-ab+b^2}{\sqrt{a^2+5ab+b^2}}\right)$
- A Cyclic Inequality in Three Variables XXII $\left(\displaystyle \sum_{cycl}\frac{a^3}{a^2+ab+b^2}\ge\frac{a+b+c}{3}\right)$
- A Cyclic Inequality in Three Variables XXIII $\left(\displaystyle 3(a^2+b^2+c^2)^2\ge 8abc(a+b+c)+\sum_{cycl}(a^2+b^2-c^2)^2\right)$
- A Cyclic Inequality in Three Variables XXIV $\left(\displaystyle \sum_{cycl} \frac{a^2b^2 (1+a^2)(1+b^2)}{(1+a)(1+b)}\geq 4(3-2\sqrt{2})abc(a+b+c)\right)$
- A Cyclic Inequality in Three Variables XXV $\left(\displaystyle \sum_{cycl} (a-\sqrt{ab}+b)^2\cdot\sum_{cycl}(a^2-ab+b^2)^2\ge 9a^2b^2c^2\right)$
- A Cyclic Inequality in Three Variables And One More $\left(\displaystyle \left(\sum_{cycl}x^{2m+2}\right)\cdot\left(\sum_{cycl}\frac{1}{(x+y)^{2m+2}}\right)\ge\frac{9}{4^{m+1}}\right)$
- Dorin Marghidanu's Cyclic Inequality in Three Variables $\left(\displaystyle\sum_{cycl}\sqrt{a^2-ab+b^2}\sqrt{b^2-bc+c^2}\ge a^2+b^2+c^2\right)$
- Dorin Marghidanu's Cyclic Inequality in Three Variables II
$\left(\displaystyle\sum_{cycl}\frac{ab}{(a+c)(b+c)} \ge\frac{3}{4}\right)$
Dorin Marghidanu's Cyclic Inequality in Three Variables III
$\left(\displaystyle \frac{a^3}{b^2+c}+\frac{b^3}{c^2+a}+\frac{c^3}{a^2+b}\gt \frac{3}{2}(abc)^{\frac{2}{3}}\right)$
- Leo Giugiuc's Cyclic Inequality in Three Variables $\left(\displaystyle a^2+b^2+c^2\ge 3\sqrt[3]{\frac{1}{4}(a-b)^2(b-c)^2(c-a)^2}+ab+bc+ca\right)$
- Tran Hoang Nam's Cyclic Inequality in Three Variables $\left(\displaystyle\sum_{cycl}(a-b)^3(a-c)^3 \le \left(\sum_{cycl}a^2-\sum_{cycl}ab\right)^3\right)$
- Cyclic Inequality with Square Roots $\left(\displaystyle 2\sqrt{2}\sum_{cycl}xy\ge\sqrt{2xyz}\sum_{cycl}\sqrt{x}+\sum_{cycl}\sqrt{x^2z^2+y^2z^2}\right)$
- Cyclic Inequality with Logarithms $\left(\displaystyle \ln \left(a^b\cdot b^c\cdot c^a\right)+6\sum_{cycl}\frac{b(1+2a)}{1+4a+a^2}\ge 3(a+b+c)\right)$
- A Cyclic Inequality with Many Sums $\left(\displaystyle\small{ \left(\sum_{cycl}a^4\right)\left(\sum_{cycl}\frac{a}{b}\right)\left(\sum_{cycl}a^3\right)\left(\sum_{cycl}\frac{a}{c}\right)\left(\sum_{cycl}a^2\right) \ge \left(\sum_{cycl}a\right)^3\left(\sum_{cycl}\frac{1}{a}\right)^2}\right)$
- Dan Sitaru's Cyclic Inequality in Three Variables III $\left(\displaystyle 4\sqrt{\sum_{cycl}\frac{a}{(a-1)^2}}\ge\sqrt{6}(10-a-b-c)\right)$
- Imad Zak's Cyclic Inequality in Three Variables $\left(\displaystyle \frac{(a+b+c)^2}{ab+bc+ca} + \frac{ab+bc+ca}{a^2+b^2+c^2}\ge 4\right)$
- Imad Zak's Cyclic Inequality in Three Variables II $\left(\displaystyle \frac{ab+bc+ca}{(a+b+c)^2} + \frac{a^2+b^2+c^2}{ab+bc+ca}\ge \frac{4}{3}\right)$
- A Simple Cyclic Inequality in Three Variables $\left(\displaystyle 3\left(\sum_{cycl}a^2\right)^2\ge 8abc(a+b+c)+\sum_{cycl}(a^2+b^2-c^2)^2\right)$
- An Inequality with Cyclic Sums on Both Sides $\left(\displaystyle \sum_{cycl}\frac{a^9}{b^6c^2}\ge\sum_{cycl}\sqrt[6]{\frac{a^{28}}{b^{17}c^5}}\right)$
- An Inequality with Cyclic Sums on Both Sides II $\left(\displaystyle \sum_{cycl}\sqrt[6]{ab^2c^3}\ge\sum_{cycl}\sqrt[30]{a^{9}b^{10}c^{11}}\right)$
- An Inequality with Cyclic Sums on Both Sides III $\left(\displaystyle \frac{x^6z^3+y^6x^3+z^6y^3}{x^2y^2z^2}\geq \frac{x^3+y^3+z^3+3xyz}{2}\right)$
- A Cyclic Inequality in Three Variables with a Variable Hierarchy $\left(\displaystyle 2(x^2y + y^2z + z^2x+xyz) \ge (x+y) (y+z) (z+x)\right)$
- Dan Sitaru's Cyclic Inequality In Three Variables $\left(\displaystyle \frac{(5a+b)(5b+c)(5c+a)}{27(a+8c)(b+8a)(c+8b)}\geq \frac{8abc}{(5a+4b)(5b+4c)(5c+a)}\right)$
- Dan Sitaru's Cyclic Inequality in Three Variables V $\left(\displaystyle (x+y+z)^2\le \sum_{cycl}\sqrt{(x^2+xy+y^2)(y^2+yz+z^2)}\right)$
- Dan Sitaru's Cyclic Inequality in Three Variables VII $\left(\displaystyle \frac{5x+3y+z}{5z+3y+x}+\frac{5y+3z+x}{5x+3z+y}+\frac{5z+3x+y}{5y+3x+z}\ge 3\right)$
- Problem 11867 from the American Mathematical Monthly $\displaystyle \left(\left(\frac{a^2}{a^2-ab+b^2}\right)^{\frac{1}{4}} + \left(\frac{b^2}{b^2-bc+c^2}\right)^{\frac{1}{4}} + \left(\frac{c^2}{c^2-ca+a^2}\right)^{\frac{1}{4}} \le 3\right)$
- An Inequality from the 1967 IMO Shortlist $\left(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{a^8+b^8+c^8}{a^3b^3c^3}\right)$
- Birth of an Inequality $\displaystyle\left(3(a^2+b^2+c^2)^2\ge 24abc\sqrt[3]{abc}+\sum_{cyc}(a^2+b^2-c^2)^2\right)$
- A Simple Inequality in Three Variables $\displaystyle\left(\sum_{cyc}ac\left(\frac{1}{2a+b}+\frac{1}{2c+b}\right)\le \sqrt{3(a^2+b^2+c^2)}\right)$
- Cyclic Inequality in Three Variables by Marian Cucoanes $\displaystyle\left( \prod_{cycl}(\sqrt{(a+b)(a+c)}-\sqrt{bc})\ge abc\right)$
- Hung Viet's Inequality IV $\left(\displaystyle \sum_{cycl}\frac{1}{a+5b}\ge\sum_{cycl}\frac{1}{a+2b+3c}\right)$
- Cyclic Inequality with Arctangents $\left(\displaystyle 4\sum_{cycl}ab\cdot\arctan\frac{c}{b}\le\pi\sum_{cycl}a^2\right)$
- A Cyclic Inequality with Powers 2 through 7 $\left(\displaystyle \sum_{cycl}\frac{(a^7+b^7)^3}{(a^4+b^4)(a^5+b^5)(a^6+b^6)}\ge 3a^2b^2c^2\right)$
- A Long Cyclic Inequality of Degree 4 $\left(\displaystyle 4\cdot\sum_{cycl}ab\cdot\sum_{cycl}a-\left(\sum_{cycl}a\right)^3\ge\frac{\displaystyle 3\sum_{cycl}ab\left[4\sum_{cycl}ab-\left(\sum_{cycl}a\right)^2\right]}{\displaystyle \sum_{cycl}a}\right)$
- A Cyclic Inequality from the 6th IMO, 1964 $\left(\displaystyle \sum_{cycl}a^2(b+c-a)\le 3abc\right)$
- Inequality from Math Phenomenon $\left(\displaystyle \frac{a^{2n+1}}{\sqrt{bc}}+\frac{b^{2n+1}}{\sqrt{ac}}+\frac{c^{2n+1}}{\sqrt{ab}}\ge a^{2n}+b^{2n}+c^{2n}\right)$
- A Cyclic Inequality from Math Phenomenon $\left(\displaystyle \frac{1}{a(b+c)}+\frac{1}{b(c+a)}+\frac{1}{c(a+b)}\ge\frac{12}{(a+b+c)^2}\right)$
- Trigonometric Inequality with Integrals $\left(\small{\Omega(a,b)+\Omega(b,c)+\Omega(c,a)\le \sqrt{2}(a^2+b^2+c^2),\;\Omega(a,b)=\int_a^{2a}\int_b^{2b}|\sin (x-y)\cos (x+y)-\sin (x+y)|dxdy}\right)$
- A Cyclic Inequality in Three Variables by Uche E. Okeke $\left(\displaystyle \frac{(a+b)(b+c)(c+a)}{2}\ge abc+ \frac{(ab+bc+ca)^2}{a+b+c}\right)$
- Problem 4142 From Crux Mathematicorum $\displaystyle\left(\Bigr(1+\frac{a^2+b^2+c^2}{ab+bc+ca}\Bigr)^{\frac{(a+b+c)^2}{a^2+b^2+c^2}}\leq \Bigr(1+\frac{a}{b}\Bigr)\Bigr(1+\frac{b}{c}\Bigr)\Bigr(1+\frac{c}{a}\Bigr)\right)$
- Two-Sided Inequality by Dorin Marghidanu $\left(\displaystyle \sqrt{2}\le\sum_{cycl}\frac{b+c}{a+\sqrt{2(b^2+c^2)}}\le 2\right)$
- A Cyclic Inequality In Three Variables by Sorin Radulescu $\left(\displaystyle \left[\sum_{cycl}x(y-z)^2\right]^3\ge 54\prod_{cycl}x(y-z)^2\right)$
- Problem 1 from the 2017 Canada MO $\left(\displaystyle\left(\frac{a}{b-c}\right)^2+\left(\frac{b}{c-a}\right)^2+\left(\frac{c}{a-b}\right)^2\gt 2\right)$
- An Easy Cyclic Inequality And a Remark $\left(\displaystyle 1\le\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\le 2\right)$
- A Cyclic Inequality from India In Three Variables And More $\left(\displaystyle \sum_{cycl}\sqrt{a^4+a^2b^2+b^4}+(6-\sqrt{3})\sum_{cycl}ab\ge 2\left(\sum_{cycl}a\right)^2\right)$
- Dan Sitaru's Cyclic Inequality in Three Variables VIII $\left(\displaystyle (xy+yz+zx)\sum_{cycl}\sqrt{x^2+xy+y^2}\le 3\sqrt{\prod_{cycl}(x^2+xy+y^2)}\right)$
- Hadamard's Determinant Inequalities and Applications I $\left((2-a-b-c+abc)^2\le (a^2+2)(b^2+2)(c^2+2)\right)$
- An Inequality with Two Triples of Variables II $\left(\displaystyle\small{ ax+by+cz+\sqrt{\left(\sum_{cycl}a^2\right)\left(\sum_{cycl}x^2\right)}\ge\frac{2}{3}\left(\sum_{cycl}a\right)\left(\sum_{cycl}x\right)}\right)$
- Two Cyclic Inequalities $\left(\displaystyle \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\ge\frac{3}{1+abc}\right),$ $\left(\begin{align} \displaystyle 3+(A+M+S)+\bigg(\frac{1}{A}+\frac{1}{M}+\frac{1}{S}\bigg)&+\bigg(\frac{A}{M}+\frac{M}{S}+\frac{S}{A}\bigg)\\&\ge\frac{3(A+1)(M+1)(S+1)}{AMS+1}\end{align}\right)$
- Ji Chen's Inequality $\left(\displaystyle (xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\ge\frac{9}{4}\right)$
- A Cyclic Inequality of Degree Four $\left(\displaystyle a^4b+b^4c+c^4a+2(a+b+c)\ge \sqrt{3}(ab+bc+ca)\right)$
- A Little More of Algebra for an Inequality, A Little Less of Calculus for a Generalization $\left(\displaystyle \sum_{cycl}(a-b)\cdot\left(\frac{a}{b}\right)^n\ge \sum_{cycl}(a-b)\cdot\frac{a}{b}\right)$
- A Cyclic Inequality in Three Variables XXVI $\left(\displaystyle \sum_{cycl}\frac{(x+y)^4+1}{(x+y)^6+1}\le\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\right)$
- Dorin Marghidanu's Inequality with Powers and Reciprocals $\left(\displaystyle \sum_{cycl}\frac{a}{a^2bc+b^4+c^4}\le\frac{1}{abc}\right)$
|Contact| |Front page| |Contents| |Algebra|
Copyright © 1996-2018 Alexander Bogomolny72265091