# Dan Sitaru's Cyclic Inequality in Three Variables VI

### Solution 1

By the Cauchy-Schwarz inequality,

$\displaystyle \left(\sqrt{a(a+2b)}+\sqrt{b(b+2a)}\right)^2\le (a+b)(a+2b+b+2a)=3(a+b)^2.$

It follows that

\displaystyle \begin{align} \sum_{cycl}\left[\sqrt{a(a+2b)}+\sqrt{b(b+2a)}\,\right]&\le \sqrt{3}\sum_{cycl}(a+b)\\ &=2\sqrt{3}\sum_{cycl}a=6\sqrt{3}. \end{align}

### Solution 2

By the Cauchy-Schwarz inequality,

\displaystyle\begin{align}\left(1\cdot\sqrt{a(a+2b)}+1\cdot\sqrt{b(b+2a)}\right)^2&\le 6\sum_{cycl}[a(a+2b)+b(b+2a)]\\ &=6\sum_{cycl}(a^2+b^2+4ab)\\ &=12(a+b+c)^2=108. \end{align}

It follows that

$\displaystyle \sum_{cycl}\left[\sqrt{a(a+2b)}+\sqrt{b(b+2a)}\,\right]\le \sqrt{108}=6\sqrt{3}.$

### Solution 3

The concavity of the square root implies (via Jensen's inequality)

\displaystyle \begin{align} LHS&\leq 6\sqrt{\frac{\displaystyle \sum_{cyc}[a(a+2b)+b(b+2a)]}{\displaystyle 6}} \\ &= 6\sqrt{\frac{2(a+b+c)^2}{6}}=6\sqrt{3}. \end{align}

### Acknowledgment

Dan Sitaru has kindly posted the problem at the CutTheKnotMath facebook page, with solutions by Abdur Rahman (Solution 1) and Serban George Florin (Solution 2). Solution 3 is by Amit Itagi. The problem was originally publshed at the Romanian Mathematical Magazine.