A Cyclic Inequality in Three Variables VI

Problem

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.

 

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