Vasile Cirtoaje's Cyclic Inequality with Three Variables

Problem

Solution 1

If one of $a,b,c$ is $0,$ the result is immediate from the AM-GM inequality. So assume all three are positive.

By the AM-GM inequality, $\displaystyle \sqrt{\frac{a+b}{c}}\le\frac{\frac{a+b}{c}+1}{2},$ so that $\displaystyle \sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}.$ Similarly, $\displaystyle \sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}$ and $\displaystyle \sqrt{\frac{b}{c+a}}\ge\frac{2b}{a+b+c}.$ Adding the three delivers the required result.

Note that equality is attained for $\displaystyle (a,b,c)=\left(\frac{1}{2},\frac{1}{2},0\right)$ and permutations.

Solution 2

We rewrite the inequality as

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

and try to prove

$\displaystyle\begin{align} \sqrt{\frac{a}{b+c}}&\ge 2\frac{a}{a+b+c},\\ \sqrt{\frac{b}{c+a}}&\ge 2\frac{b}{a+b+c},\\ \sqrt{\frac{c}{a+b}}&\ge 2\frac{c}{a+b+c}. \end{align}$

Suffice it to consider the first one. We modify:

$\displaystyle\begin{align} &\sqrt{\frac{a}{b+c}}\ge 2\frac{a}{a+b+c}&~\Leftrightarrow\\ &\sqrt{\frac{a}{1-a}}\ge 2\frac{a}{a+1-a}&~\Leftrightarrow\\ &\sqrt{\frac{a}{1-a}}\ge 2a&~\Leftrightarrow\\ &\frac{a}{1-a}\ge 4a^2&~\Leftrightarrow\\ &\frac{1}{1-a}\ge 4a&~\Leftrightarrow\\ & 4a^2-4a+1\ge 0&~\Leftrightarrow\\ & (2a-1)^2\ge 0, \end{align}$

which is true.

Acknowledgment

Leo Giugiuc has kindly communicated to me the above statement that he used to solve another problem. Solution 1 is Leo's; he credits Vasile Cirtoaje for the statement above.

Solution 2 is by Nezir Hotic.

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