# A Cyclic Inequality in Three Variables VII

### Problem

### Solution 1

First off, observe that $\sqrt{2(u^2+v^2)}=\sqrt{(u+v)^2+(u-v)^2}\ge u+v.$ It follows that

$\displaystyle\begin{align} \sum_{cycl}x\sqrt{x^2z^2+y^4} &\ge \frac{1}{\sqrt{2}}\sum_{cycl}(x^2z+xy^2)\\ &\ge\frac{1}{\sqrt{2}}\sum_{cycl}2xy\sqrt{xz}\\ &\ge\sqrt{2}\sum_{cycl}xz\sqrt{yz}. \end{align}$

### Solution 2

By the AM-GM inequality,

$\displaystyle\begin{align} \sum_{cycl}x\sqrt{x^2z^2+y^4} &\ge \sum_{cycl}x\sqrt{2\sqrt{(x^2z^2)(y^4)}}\\ &\ge \sqrt{2}\sum_{cycl}xy\sqrt{xz}\\ &\ge\sqrt{2}\sum_{cycl}xy\sqrt{xz}. \end{align}$

Equality is achieved for $x=y=z.$

### Acknowledgment

Dan Sitaru has kindly posted the above problem (from his book "Math Accent") at the CutTheKnotMath facebook page. Solution 1 is by Ravi Prakash.

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