# A Problem from the Danubius-XI Contest

### Solution 1

First, we prove

Lemma

If $x,y\gt 0,\,$ then

$\displaystyle \frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}\ge\frac{1}{1+xy}.$

Indeed, $\displaystyle (x+y)\left(\frac{1}{x}+y\right)\ge (1+y)^2;\,$ $\displaystyle (y+x)\left(\frac{1}{y}+x\right)\ge (1+x)^2.\,$ Equality is achieved if $\displaystyle \frac{x}{1/x}=\frac{y}{y},\,$ i.e., $x^2=1,\,$ or $x=1\,$ and, similarly, if $\displaystyle \frac{y}{1/y}=\frac{x}{x},\,$ i.e., $y^2=1,\,$ or $y=1.\,$ Hence,

\displaystyle \begin{align} \frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}&\ge\frac{\displaystyle 1}{\displaystyle (x+y)\left(\frac{1}{x}+y\right)}+\frac{\displaystyle 1}{\displaystyle (y+x)\left(\frac{1}{y}+x\right)}\\ &=\frac{1}{x+y}\left(\frac{y}{1+xy}+\frac{x}{1+xy}\right)=\frac{1}{1+xy}. \end{align}

The lemma implies

\displaystyle \begin{align} &\frac{a}{(1+b)^2}+\frac{a}{(1+c)^2}\ge\frac{a}{1+bc},\\ &\frac{b}{(1+c)^2}+\frac{b}{(1+a)^2}\ge\frac{b}{1+ca},\\ &\frac{c}{(1+a)^2}+\frac{c}{(1+b)^2}\ge\frac{c}{1+ab}. \end{align}

Adding up and using Bergström's inequality we get

\displaystyle \begin{align} \frac{a+b}{(1+c)^2}+\frac{b+c}{(1+a)^2}+\frac{c+a}{(1+b)^2}&\ge\frac{a}{1+bc}+\frac{b}{1+ca}+\frac{c}{1+ab}\\ &=\frac{a^2}{a+abc}+\frac{b^2}{b+abc}+\frac{c^2}{c+abc}\\ &\ge\frac{(a+b+c)^2}{a+b+c+3abc}=\frac{S^2}{S+3abc}\\ &\ge\frac{\displaystyle S^2}{\displaystyle S+3\left(\frac{S}{3}\right)^3}=\frac{9S}{9+S^2}. \end{align}

Equality occurs for $a=b=c=1.$

### Solution 2

The inequality is equivalent to $\displaystyle \sum_{cycl}\frac{S-c}{(1+c)^2}\ge \frac{9S}{9+S^2}.$

Consider $\displaystyle f(x)=\frac{S-x}{(1+x)^2}.\,$ $\displaystyle f''(x)=\frac{6S-2x+4}{(x+1)^4},\,$ so that $f''(x)\ge 0,\,$ for $x\le 3S+2\,$ which holds for $x\le S.$

$\displaystyle \sum_{cycl}\frac{S-c}{(1+c)^2}\ge 3f\left(\frac{S}{3}\right)=\frac{\displaystyle 2S}{\displaystyle \left(1+\frac{S}{3}\right)^2}.\,$ Thus, suffice it to show that

$\displaystyle \frac{\displaystyle 2S}{\displaystyle \left(1+\frac{S}{3}\right)^2}\ge\frac{9S}{9+S^2}.$

But this is equivalent to $9+S^2-6S\ge 0,\,$ or $(3-S)^2\ge 0\,$ which is true. Equality for $S=3.$

### Solution 3

By Hölder's inequality,

$\displaystyle \left(\sum_{cycl}\frac{a+b}{(1+c)^2}\right)\left(\sum_{cycl}(a+b)(1+c)\right)^2\ge 8S^2.$

Further

$\displaystyle \sum_{cycl}(a+b)(1+c)=2\sum_{cycl}a+2\sum_{cycl}ab\le 2S+\frac{2S^2}{3}.$

Thus, applying the Cauchy-Schwarz inequality,

\displaystyle \begin{align} \sum_{cycl}\frac{a+b}{(1+c)^2}&\ge\frac{\displaystyle 8S^3}{\displaystyle \left(\sum_{cycl}(a+b)(a+c)\right)^2}\\ &\ge\frac{\displaystyle 8S^3}{\displaystyle 4S^2\left(1+\frac{S}{3}\right)^2}\\ &\ge\frac{\displaystyle 8S^3}{\displaystyle 4S^2\cdot 2\left(1+\frac{S^2}{9}\right)}=\frac{9S}{9+S^2}. \end{align}

### Solution 4

We'll use weighted Jensen's inequality for the convex function $\displaystyle f(x)=\frac{1}{(1+x)^2}.\,$ Note that indeed $\displaystyle f''(x)=\frac{6}{(1+x)^4}\ge 0.$

\displaystyle \begin{align} \sum_{cycl}\frac{a+b}{(1+c)^2} &= \sum_{cycl}(a+b)f(c)\\ &\ge\sum_{cycl}(a+b)f\left(\frac{\displaystyle \sum_{cycl}(a+b)c}{\displaystyle \sum_{cycl}(a+b)}\right)\\ &=2S\frac{\displaystyle 1}{\displaystyle \left[1+\frac{2(ab+bc+ca)}{2(a+b+c)}\right]^2}\ge\frac{\displaystyle 2S}{\displaystyle \left(1+\frac{S}{3}\right)^2}\\ &=\frac{18S}{(3+S)^2}\ge\frac{9S}{9+S^2}. \end{align}

### Acknowledgment

The above problem has been kindly posted at the CutTheKnotMath facebook page by Dorin Marghidanu. Solution 1 is by Dorin Marghidanu; Solution 2 is by Imad Zak; Solution 3 is by Kevin Soto Palacios; Solution 4 is by Marian Dinca.

Copyright © 1996-2018 Alexander Bogomolny

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