# Problem 11867 from the American Mathematical Monthly

### Solution

Since $-xy\ge -|xy|\,$ implies $\displaystyle \frac{x^2}{x^2-xy+y^2}\le\frac{x^2}{x^2-|xy|+y^2},\,$ suffice it consider positive $a,b,c.$

Lemma

Fro $x\gt 0,\,$ $\displaystyle\frac{1}{1-x+x^2}\le\frac{4}{(x+1)^2}.$

This is equivalent to $0\le 3(x-1)^2,\,$ which is obviously true.

From the lemma, with $\displaystyle x=\frac{a}{b},\,$ $\displaystyle \left(\frac{a^2}{a^2-ab+b^2}\right)^{\frac{1}{4}}\le\sqrt[4]{\frac{4a^2}{(a+b)^2}}=\sqrt{\frac{2a}{a+b}}.\,$ Similarly, $\displaystyle \left(\frac{b^2}{b^2-bc+c^2}\right)^{\frac{1}{4}}\le\sqrt{\frac{2b}{b+c}}\,$ and $\displaystyle \left(\frac{c^2}{c^2-ca+a^2}\right)^{\frac{1}{4}}\le\sqrt{\frac{2c}{c+a}}.\,$ Suffice it to prove that

$\displaystyle \sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{c+a}}\le 3.$

Note that $\displaystyle \sqrt{\frac{2a}{a+b}}=\sqrt{\frac{2a}{(a+b)(c+a)}}\cdot\sqrt{c+a}\,$ and, similarly, $\displaystyle \sqrt{\frac{2b}{b+c}}=\sqrt{\frac{2b}{(b+c)(a+b)}}\cdot\sqrt{a+b}\,$ and $\displaystyle \sqrt{\frac{2c}{c+a}}=\sqrt{\frac{2c}{(c+a)(b+c)}}\cdot\sqrt{b+c}.\,$ We apply the Cauchy-Schwarz inequality:

\displaystyle \begin{align} \sum_{cycl}\sqrt{\frac{2a}{(a+b)(c+a)}}\cdot\sqrt{c+a}&\le\sqrt{\sum_{cycl}\frac{2a}{(a+b)(c+a)}\cdot\sum_{cycl}(a+b)}\\ &=\sqrt{\frac{8(a+b+c)(ab+bc+ca)}{(a+b)(b+c)(c+a)}}. \end{align}

Thus, suffice it to show that $\displaystyle \sqrt{\frac{8(a+b+c)(ab+bc+ca)}{(a+b)(b+c)(c+a)}}\le 3,\,$ i.e.,

$8(a+b+c)(ab+bc+ca)\le 9[(a+b+c)(ab+bc+ca)-abc]$

which is the same as $9abc\le(a+b+c)(ab+bc+ca).\,$ The latter is true by the AM-GM inequality.

### Illustration

With the substitution $\displaystyle x=\frac{b}{a},\,y=\frac{b}{c},\,z=\frac{c}{a},\,$ the problem reduces to proving

$\displaystyle \left(\frac{1}{1-x+x^2}\right)^{\frac{1}{4}}+\left(\frac{1}{1-y+y^2}\right)^{\frac{1}{4}}+\left(\frac{1}{1-z+z^2}\right)^{\frac{1}{4}}\le 3,$

subject to $xyz=1.\,$

Since the function $f(t)=t^{\frac{1}{4}}\,$ is concave, by Jensen's inequality,

$\displaystyle \frac{1}{3}\sum_{cycl}\left(\frac{1}{1-x+x^2}\right)^{\frac{1}{4}}\le\left(\frac{1}{3}\sum_{cycl}\frac{1}{1-x+x^2}\right)^{\frac{1}{4}}.$

Thus suffice it to prove that

$\displaystyle 3\left(\frac{1}{3}\sum_{cycl}\frac{1}{1-x+x^2}\right)^{\frac{1}{4}}\le 3$

which is equivalent to

$\displaystyle \sum_{cycl}\frac{1}{1-x+x^2}=\sum_{cycl}\frac{1+x}{1+x^3}\le 3.$

Since $xyz=1,\,$ this reduces to

$\displaystyle f(x,y)=\frac{1+x}{1+x^3}+\frac{1+y}{1+y^3}+\frac{(xy)^2(1+xy)}{1+(xy)^3} \le 3.$

The graph of the function $f\,$ appears below.

### Acknowledgment

This is problem 11867 from the American Mathematical Monthly. The problem has been proposed by George Apostolopoulos. The solution is by Leo Giuguic.