# Dan Sitaru's Cyclic Inequality In Three Variables with Constraints III

### Problem

### Solution 1

Let $x=a^2,\,$ $b=y^2,\,$ $c=z^2,\,$ where $a,b,c\gt 0.\,$ Then $ab+bc+ca=2.\,$ Now consider,

$\displaystyle\begin{align} &\frac{a^3}{b}+\frac{b^3}{a}+6ab-4a^2-4b^2\\ &\qquad\qquad=\frac{1}{ab}(a^4-4a^3b+6a^2b^2-4ab^3+b^4)\\ &\qquad\qquad=\frac{1}{ab}(a-b)^4\ge 0. \end{align}$

By deriving two similar inequalities and summing up we get

$\displaystyle 6\sum_{cycl}ab+\sum_{cycl}\left(\frac{a^3}{b}+\frac{b^3}{a}\right)\ge 8\sum_{cycl}a^2$

which is, after a face lift, the required inequality.

### Solution 2

$\displaystyle 6\sum_{cycl}\sqrt{xy}+\sum_{cycl}\left(\sqrt{\frac{x^3}{y}}+\sqrt{\frac{x^3}{y}}\right)\ge 4\sum_{cycl}(x+y).$

Suffice it to prove that $\displaystyle (\sqrt{\frac{x^3}{y}}+\sqrt{\frac{x^3}{y}}+6\sqrt{xy}\ge 4(x+y)\,$ which is equivalent to $x^2+y^2+6xy\ge 4\sqrt{xy}(x+y),\,$ i.e. $(\sqrt{x}-\sqrt{y})^4\ge 0.$

Summing up yields the required inequality. Equality is achieved for $\displaystyle x=y=z=\frac{2}{3}.$

### Acknowledgment

Dan Sitaru has kindly posted a problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page. Solution 1 is by Ravi Prakash (India); Solution 2 is by Kevin Soto Palacios (Peru). Hoang Tung, Aziz, Sanong Hauerai have independently arrived at variants of those.

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