Cyclic Inequality with Logarithms


Cyclic Inequality with Logarithms

Solution 1

First we prove


For $a,b\ge 1,$

$\displaystyle b\ln a+\frac{6b(1+2a)}{a^2+4a+1}\ge 3b.$

Indeed, set $f(b)=\displaystyle b\ln a+\frac{6b(1+2a)}{a^2+4a+1}-3b\,$ on $[1,\infty).\,$ $\displaystyle f'(b)=\ln a+\frac{6(1+2a)}{a^2+4a+1}-3.\,$

Now let $g(a)=(a^2+4a+1)f'(b)=(a^2+4a+1)\ln a-3a^2+3,\,$ on $[1,\infty).],$ We have

$\displaystyle\begin{align}g'(a)&=2(a+2)\ln a-5a+4+\frac{1}{a},\\ g''(a)&=2\ln a-3+\frac{4}{a}-\frac{1}{a^2},\\ g'''(a)&=\frac{2(a-1)^2}{a^3}\ge 0,\;a\ge 1. \end{align}$

We deduce that $g''(a),g'(a),g(a)\,$ are all increasing for $a\ge 1,\,$ implying that so is $f\,$ and since $f(1)=\displaystyle\frac{g(a)}{a^2+4a+1}\ge 0,\,$ the conclusion follows.

The other two inequalities are treated in a similar manner and then added to obtain the required inequality.

$f'(x)\lt 0,\;$ for $x\lt 1,\;$ and $f'(x)\gt 0,\;$ for $x\gt 1.\;$ Since $f(1)=0,\;$ $f(x)\ge 0,\;$ for $x\gt 0.$

Solution 2

First we prove


Function $\displaystyle f(x)=\ln x+\frac{6(1+2x)}{x^2+4x+1}\,$ is strictly increasing on $[0,\infty).$

Indeed, $\displaystyle f'(x)=\frac{1}{x}-\frac{12(x^2+x+1)}{(x^2+4x+1)^2}.\,$ Further,

$f'(x)\ge 0\Longleftrightarrow[(x^2+x+1)+3x]^2\ge 12x(x^2+x+1),$

which is true by the AM-GM inequality for $u=x^2+x+1\,$ and $v=3x.$

Back to the problem: by the lemma, $f(x)\ge f(1)=3,\,$ $x\ge 1.\,$ Thus, $bf(a)+cf(b)+af(c)\ge 3(a+b+c),\,$ implying the required inequality.


The problem above has been posted on the CutTheKnotMath facebook page by Dan Sitaru. Leo Giugiuc submitted two solutions (Solution 1 and Solution 2); Diego Alvariz submitted a solution, practically the same as Solution 2.


Cyclic inequalities in three variables

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny