# A Cyclic Inequality in Three Variables III

### Solution 1

Denote $a+b+c=x.\;$ Using the AM-GM,

$\displaystyle\sqrt{\frac{b+c}{a}}\le\frac{\displaystyle\frac{b+c}{a}+1}{2}=\frac{x}{2a},$

implying $\displaystyle\sqrt{\frac{a}{b+c}}\ge\frac{2a}{x}.\,$ By rotation of variables be obtain two additional inequalities which leads to

$\displaystyle\sum_{cycl}\displaystyle\sqrt{\frac{a}{b+c}}\ge\sum_{cycl}\frac{2a}{x}=\frac{2a+2b+2c}{a+b+c}=2.$

For equality we'd need $\displaystyle\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=1,\,$ which is impossible. It follows that

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

### Solution 2

Using the GM-HM inequality,

\displaystyle\begin{align} \sum_{cycl}\sqrt{\frac{a}{b+c}} &= \sum_{cycl}\sqrt{\frac{a(b+c)}{(b+c)^2}}=\sum_{cycl}\frac{\sqrt{a(b+c)}}{b+c}\\ &\ge\sum_{cycl}\frac{\displaystyle\frac{2}{\displaystyle\frac{1}{a}+\frac{1}{b+c}}}{b+c}=\sum_{cycl}\frac{2a(b+c)}{(a+b+c)(b+c)}\\ &=\sum_{cycl}\frac{2a}{a+b+c}\\ &=2. \end{align}

### Solution 3

\displaystyle\begin{align} \sum_{cycl}\sqrt{\frac{a}{b+c}} &= \sum_{cycl}\sqrt{\frac{a^2}{a(b+c)}}=\sum_{cycl}\frac{a}{\sqrt{a(b+c)}}\\ &\ge\sum_{cycl}\frac{a}{\displaystyle \frac{a+b+c}{2}}=\sum_{cycl}\frac{2a}{a+b+c}\\ &=2. \end{align}

### Solution 4

Function $y=\sqrt{x}\,$ is concave in the domain of its definition, $x\ge 0.\,$ By Jensen's inequality, then

$\displaystyle\frac{f(x_1)+f(x_2)+f(x_3)}{3}\ge f\left(\frac{x_1+x_2+x_3}{3}\right).$

With $\displaystyle x_1=\frac{a}{b+c},\,$ $\displaystyle x_2=\frac{b}{c+a},\,$ $\displaystyle x_3=\frac{c}{a+b},\,$ this translates, via Nesbitt's inequality, into

\displaystyle\begin{align} \sum_{cycl}\sqrt{\frac{a}{b+c}} &\ge 3\sqrt{\frac{\displaystyle\sum_{cycl}\frac{a}{b+c}}{3}}\\ &\ge 3\sqrt{\frac{\displaystyle\frac{3}{2}}{3}}=\frac{3}{2}\sqrt{2}\\ &\gt 2. \end{align}

The above derivation lends itself to a simple generalization:

Let $a,b,c\gt 0\,$ and $n\gt 1.\,$ Then

$\displaystyle\sqrt[n]{\frac{a}{b+c}}+\sqrt[n]{\frac{b}{c+a}}+\sqrt[n]{\frac{c}{a+b}}\ge \frac{3}{2}\sqrt[n]{2^{n-1}}.$

Equality is only achieved for $a=b=c.$

### Acknowledgment

I have borrowed the above problem from the Romanian Mathematical Magazine where it appeared under the caption "Russian Inequality - 2" with three solution: Solution 1 by Rovsen Pirkuliyev (Azerbaidjian); Solution 2 by Myagmarsuren Yadamsuren (Mongolia); Solution 3 by Marian Dinca (Romania). The inequality is weak. Proof 4 highlights the way in which it can be improved and generalized.