# 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.