Simple Inequality with a Variety of Solutions

Problem

Simple Inequality with a Variety of Solutions

Solution 1

Let $a=\ln x,\,$ $b=\ln y,\,$ $c=\ln z.\,$ The inequality becomes

$\displaystyle \left(\frac{a}{bc}+\frac{b}{ca}\right)+\left(\frac{b}{ca}+\frac{c}{ab}\right)+\left(\frac{c}{ab}+\frac{a}{bc}\right)\ge\frac{18}{a+b+c},$

or, equivalently,

$\displaystyle \left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)(a+b+c)\ge 9.$

This is seen to be true by first applying the Cauchy-Schwarz and then the AM-GM inequality:

$\displaystyle\begin{align} \left(\sum_{cycl}\frac{a}{bc}\right)(a+b+c)&\ge \frac{(a+b+c)^2}{3abc}\cdot (a+b+c)\\ &=\frac{(a+b+c)^3}{3abc}\ge 9. \end{align}$

Solution 2

Let $a=\ln x,\,$ $b=\ln y,\,$ $c=\ln z.\,$ The inequality reduces to

$\displaystyle \left(\frac{a}{bc}+\frac{b}{ca}\right)+\left(\frac{b}{ca}+\frac{c}{ab}\right)+\left(\frac{c}{ab}+\frac{a}{bc}\right)\ge\frac{18}{a+b+c},$

$\displaystyle \begin{align} LHS&=\sum_{cycl}\left(\frac{a}{bc}+\frac{b}{ca}\right)=\sum_{cycl}\frac{a^2+b^2}{abc}\\ &\ge\sum_{cycl}\frac{2ab}{abc}=2\sum_{cycl}\frac{1}{a}\\ &\overset{AM-HM}{\ge}2\cdot\frac{9}{a+b+c}=\frac{18}{a+b+c}. \end{align}$

Solution 3

Let $a=\ln x,\,$ $b=\ln y,\,$ $c=\ln z.\,$ The inequality reduces to

$\displaystyle \left(\frac{a}{bc}+\frac{b}{ca}\right)+\left(\frac{b}{ca}+\frac{c}{ab}\right)+\left(\frac{c}{ab}+\frac{a}{bc}\right)\ge\frac{18}{a+b+c},$

$\displaystyle \begin{align} LHS&=\sum_{cycl}\left(\frac{a}{bc}+\frac{b}{ca}\right)=2\sum_{cycl}\frac{a}{bc}\\ &\ge\frac{2}{abc}\sum_{cycl}a^2\ge\frac{2}{abc}\sum_{cycl}ab\\ &=2\sum_{cycl}\frac{1}{a}\overset{AM-HM}{\ge}2\cdot\frac{9}{a+b+c}\\ &=\frac{18}{a+b+c}. \end{align}$

Solution 4

Let $a=\ln x,\,$ $b=\ln y,\,$ $c=\ln z.\,$ The inequality reduces to

$\displaystyle \left(\frac{a}{bc}+\frac{b}{ca}\right)+\left(\frac{b}{ca}+\frac{c}{ab}\right)+\left(\frac{c}{ab}+\frac{a}{bc}\right)\ge\frac{18}{a+b+c},$

$\displaystyle \begin{align} LHS&=\sum_{cycl}\left(\frac{a}{bc}+\frac{b}{ca}\right)=\sum_{cycl}\frac{a^2+b^2}{abc}\\ &\ge\sum_{cycl}\frac{2ab}{abc}=2\sum_{cycl}\frac{1}{a}\\ &=2\cdot 3\cdot\frac{\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}\overset{AM-HM}{\ge}6\cdot\frac{3}{\displaystyle \frac{1}{1/a}+\frac{1}{1/b}+\frac{1}{1/c}}\\ &=\frac{18}{a+b+c}. \end{align}$

Solution 5

$x,y,z\gt 1\,\Rightarrow\,\ln x,\,\ln y,\,\ln z\gt 0.\,$

$\displaystyle \begin{align} LHS&=\sum_{cycl}\left(\frac{\ln x}{\ln y\ln z}+\frac{\ln y}{\ln z\ln x}\right)\overset{AM-GM}{\ge} 2\left(\frac{1}{\ln x}+\frac{1}{\ln y}+\frac{1}{\ln z}\right)\\ &\overset{Cauchy-Schwarz}{\ge}2\cdot\frac{(1+1+1)^2}{\ln x+\ln y+\ln z}=\frac{18}{\ln (xyz)}. \end{align}$

Solution 6

Let $a=\ln x,\,$ $b=\ln y,\,$ $c=\ln z.\,$ The inequality reduces to

$\displaystyle \left(\frac{a}{bc}+\frac{b}{ca}\right)+\left(\frac{b}{ca}+\frac{c}{ab}\right)+\left(\frac{c}{ab}+\frac{a}{bc}\right)\ge\frac{18}{a+b+c},$

or, equivalently,

$\displaystyle \frac{1}{abc}\sum_{cycl}a^2\ge\frac{9}{a+b+c},$

or,

$\displaystyle \sum_{cycl}a^2\cdot\sum_{cycl}a\ge 9.$

By the AM-GM inequality

$\displaystyle\begin{align} &\sum_{cycl}a^2\ge 3\sqrt[3]{a^2b^2c^2}\\ &\sum_{cycl}a\ge 3\sqrt[3]{abc}\\ \end{align}$

the product of which is $\displaystyle \sum_{cycl}a^2\cdot\sum_{cycl}a\ge 9abc,\,$ which is equivalent to the required inequality.

Solution 7

Let $a=\ln x,\,$ $b=\ln y,\,$ $c=\ln z.\,$ The inequality reduces to

$\displaystyle \left(\frac{a}{bc}+\frac{b}{ca}\right)+\left(\frac{b}{ca}+\frac{c}{ab}\right)+\left(\frac{c}{ab}+\frac{a}{bc}\right)\ge\frac{18}{a+b+c},$

or, equivalently,

$\displaystyle \frac{1}{abc}\sum_{cycl}a^2\ge\frac{9}{a+b+c},$

$\displaystyle \begin{align} LHS&=\frac{1}{abc}\sum_{cycl}a^2\ge\frac{2}{abc}\frac{\displaystyle \left(\sum_{cycl}a\right)^2}{3}\\ &\overset{AM-GM}{\ge}\frac{2}{3}\frac{\displaystyle \left(\sum_{cycl}a\right)^2}{\displaystyle \left(\sum_{cycl}a\right)^3}\cdot\frac{27}{1}=\frac{18}{\displaystyle \sum_{cycl}a}. \end{align}$

Solution 8

Let $a=\ln x,\,$ $b=\ln y,\,$ $c=\ln z.\,$ The inequality reduces to

$\displaystyle \left(\frac{a}{bc}+\frac{b}{ca}\right)+\left(\frac{b}{ca}+\frac{c}{ab}\right)+\left(\frac{c}{ab}+\frac{a}{bc}\right)\ge\frac{18}{a+b+c},$

$\displaystyle \begin{align} LHS&=\sum_{cycl}\frac{a}{bc}\overset{Chebyshev}{\ge}\frac{1}{3}(a+b+c)\left(\sum_{cycl}\frac{1}{bc}\right)\\ &\overset{Cauchy-Schwarz}{\ge}\frac{3(a+b+c)}{ab+bc+ca}\ge\frac{9}{a+b+c} \end{align}$

because $(a+b+c)^2\ge 3(ab+bc+ca)\,$ which follows from $a^2+b^2+c^2\ge ab+bc+ca.$

Solution 9

Let $\ln x=t_1,\,\ln y=t_2,\,\ln z=t_3.\,$ The inequality reduces to

$\displaystyle \begin{align} LHS&=\sum_{cycl}\frac{t_1}{t_2t_3}=\sum_{cycl}\frac{1}{t_3}\left(\frac{t_1}{t_2}+\frac{t_2}{t_1}\right)\\ &\overset{AM-GM}{\ge}2\cdot\sum_{cycl}\frac{1}{t_3}\overset{Bergstrom}{\ge}2\cdot\frac{9}{\displaystyle \sum_{cycl}t_1}\\ &=\frac{18}{t_1+t_2+t_3}. \end{align}$

Solution 10

Let $a=\ln x,\,$ $b=\ln y,\,$ $c=\ln z.\,$ The inequality reduces to

$\displaystyle \sum_{cycl}\frac{a}{bc}\overset{AM-GM}{\ge}3\sqrt[3]{\frac{1}{abc}}\overset{AM-GM}{\ge}\frac{9}{a+b+c}.$

Solution 11

$\displaystyle \begin{align} LHS &= 2\sum_{cycl}\frac{\ln x}{\ln y\ln z}\ge 3\sqrt[3]{\frac{8}{\ln x\ln y\ln z}}\\ &=\frac{6}{\sqrt[3]{\ln x\ln y\ln z}}. \end{align}$

AM-GM redux:

$\displaystyle 3\sqrt[3]{\frac{8}{\ln x\ln y\ln z}}\le\ln x+\ln y+\ln z=\ln (xyz).$

Hence $\displaystyle LHS\ge\frac{18}{\ln (xyz)}.$

Acknowledgment

Dan Sitaru has kindly posted the problem from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page and later commented with several proofs. Solution 1 is by Kevin Soto Palacios; Solution 2 is by Nirapada Pal; Solution 3 is by Dan Sitaru; Solution 4 is by Nikolaos Skoutaris; Solution 5 is by Nguyen Thanh Nho; Solution 6 is by Soumava Chakraborty, Geanina Tudose came up with the same solution; Solution 7 is by Eliezer Okeke; Solution 8 is by Seyran Ibrahimov; Solution 9 is by Myagmarsuren Yadamsuren; Solution 11 is by N. N. Taleb.

 

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