An Easy Cyclic Inequality And a Remark

Problem

An Easy Cyclic Inequality And a Remark

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

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