# A Cyclic Inequality in Three Variables VI

### Solution 1

\displaystyle\begin{align} \sum_{cycl}\sqrt{\frac{a+b}{2ac}}+\sum_{cycl}\sqrt{\frac{2a}{c(a+b)}} &\le \sum_{cycl}\sqrt{\frac{(a+b)b}{2abc}}+\sum_{cycl}\sqrt{\frac{(a+b)a}{2abc}}\\ &\le\frac{\displaystyle\sum_{cycl}\sqrt{a+b}(\sqrt{a}+\sqrt{b})}{\sqrt{2abc}}\\ &\le\frac{\displaystyle\sum_{cycl}\sqrt{2}\left(\sqrt{a+b}\right)^2}{\sqrt{2abc}}.\\ \end{align}

It follows that

\displaystyle\begin{align} \sum_{cycl}\sqrt{\frac{a+b}{2ac}}+\sum_{cycl}\sqrt{\frac{2a}{c(a+b)}} &\le \frac{\displaystyle\sum_{cycl}\sqrt{2}\left(\sqrt{a+b}\right)^2}{\sqrt{2abc}}\\ &=\frac{2\sqrt{2}(a+b+c)}{\sqrt{2abc}}=\frac{2(a+b+c)}{\sqrt{abc}}, \end{align}

as required.

### Solution 2

Using the Cauchy-Schwarz inequality,

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

Again, using the Cauchy-Schwarz inequality,

\displaystyle\begin{align} \sum_{cycl}\sqrt{\frac{2a}{c(a+b)}} &\le \sqrt{2\sum_{cycl}a} \sqrt{\sum_{cycl}\frac{1}{c(a+b)}}\le \sqrt{2\sum_{cycl}a}\sqrt{\sum_{cycl}\frac{1}{c(2\sqrt{ab})}}\\ &= \sqrt{\sum_{cycl}a} \sqrt{ \frac{1}{\sqrt{abc}} \sum_{cycl}\frac{1}{\sqrt{a}}}= \sqrt{\sum_{cycl}a} \sqrt{ \frac{1}{\sqrt{abc}} \sum_{cycl}\frac{bc}{\sqrt{abc}}}\\ &= \sqrt{\frac{\displaystyle\sum_{cycl}a}{abc}} \sqrt{\sum_{cycl}\sqrt{ab}}\le \sqrt{ \frac{\displaystyle\sum_{cycl}a}{abc}} \sqrt{ \sqrt{\sum_{cycl}a} \sqrt{\sum_{cycl}b} }\\ &\le \frac{\displaystyle\sum_{cycl}a}{\sqrt{abc}}\\ \end{align}

Now it only remains to add the two ineqalities.

### Acknowledgment

Dan Sitaru has kindly posted the problem from his book, Math Accent, with two solutions, at the CutTheKnotMath facebook page. Solution 1 is by Kevin Soto Palacios; Solution 2 is by Soumava Chakraborty.