Problem 11867 from the American Mathematical Monthly
Problem
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.
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 $\left(\displaystyle\frac{10a^3}{3a^2+7bc}+\frac{10b^3}{3b^2+7ca}+\frac{10c^3}{3a^2+7ab}\ge a+b+c\right)$
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Dorin Marghidanu's Cyclic Inequality in Three Variables III
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