# Cyclic Inequality with Square Roots

### Solution 1

Set $x=a^2\,$ $y=b^2,\,$ $z=c^2.\,$ We choose $a,b,c\ge 0.\,$ The given inequality is equivalent to

$\displaystyle 2\sqrt{2}(a^2b^2+b^2c^2+c^2a^2)\ge\sqrt{2a^2b^2z}(a+b+c)+\sum_{cycl}a^2\sqrt{c^4+b^4}.$

Let's prove $c^4+b^4\le 2(b^2-bc+c^2)^2.\,$ This is equivalent to

$c^4+b^4\le2(b^2+c^2)^2+2b^2c^2-4bc(b^2+c^2)$

and, in turn, to

\displaystyle\begin{align} &(b^4+c^4+2b^2c^2)+4b^2c^2-4bc(b^2+c^2)\\ &\qquad\qquad\qquad=(b^2+c^2)^2-4bc(b^2+c^2)+4b^2c^2\\ &\qquad\qquad\qquad=(b-c)^4\ge 0. \end{align}

So, we have

\displaystyle\begin{align} \sum_{cycl}a^2\sqrt{c^4+b^4}&\le\sum_{cycl}\sqrt{2}a^2(b^2-bc+c^2)\\ &=2\sqrt{2}(a^2b^2+b^2c^2+c^2a^2)\\ &\qquad\qquad\qquad-\sqrt{2}abc(a+b+c)\,\Leftrightarrow\\ 2\sqrt{2}(a^2b^2+b^2c^2+c^2a^2)&\ge\sqrt{2}abc(a+b+c)+\sum_{cycl}a^2\sqrt{c^4+b^4}. \end{align}

### Solution 2

\displaystyle\begin{align} 2\sqrt{2}\sum_{cycl}xy &=\sqrt{2}\sum_{cycl}(xy+zx)\\ &=\sqrt{2}\sum_{cycl}x(y+z)\\ &=\sum_{cycl}x\sqrt{2(y+z)^2}\\ &=\sum_{cycl}x\sqrt{2(x^2+2xy+y^2)}\\ &=\sum_{cycl}x\sqrt{(1^2+1^2)\left[\left(\sqrt{y^2+z^2}\right)^2+(\sqrt{2yz})^2\right]}\\ &\ge\sum_{cycl}x\left(\sqrt{y^2+z^2}+\sqrt{2yz}\right),\,\text{by the Cauchy-Schwarz inequality}\\ &=\sum_{cycl}\sqrt{x^2y^2+z^2x^2}+\sum_{cycl}\sqrt{2xyz}\sqrt{x}\\ &=\sqrt{2xyz}(\sqrt{x}+\sqrt{y}+\sqrt{z})+\sum_{cycl}\sqrt{x^2y^2+y^2z^2}. \end{align}

### Acknowledgment

Dan Sitaru has kindly posted this problem from the Romanian Mathematical Magazine, with two solutions, at the CutTheKnotMath facebook page. Solution 1 is by Kevin Soto Palacios; Solution 2 is by Myagmarsuren Yadamsuren.