# Dan Sitaru's Cyclic Inequality in Three Variables

### Solution 1

\displaystyle \begin{align} &(a-b)^2\geq 0\\ &a^2-2ab+b^2\geq 0\\ &24a^2+27ab+3b^2\leq 25a^2+25ab+4b^2\\ &3(8a^2+8ab+ab+b^2)\leq 25a^5+5ab+20ab+4b^2\\ &3(8ab+b)(a+b)\leq (5a+4b)(5a+b)\\ &\frac{5a+b}{3(8a+b)}\geq \frac{a+b}{5a+4b}\overbrace{\geq}^{AM-GM}\frac{2\sqrt{ab}}{5a+4b}. \end{align}

(1)

$\displaystyle \frac{5a+b}{3(8a+b)}\geq \frac{2\sqrt{ab}}{5a+4b}.$

(2)

$\displaystyle \frac{5b+c}{3(8b+c)}\geq \frac{2\sqrt{bc}}{5b+4c}.$

(3)

$\displaystyle \frac{5c+a}{3(8c+a)}\geq \frac{2\sqrt{ca}}{5c+4a}.$

By multiplying the relationships (1), (2), (3),

$\displaystyle \frac{(5a+b)(5b+c)(5c+a)}{27(a+8c)(b+8a)(c+8b)}\geq \frac{8abc}{(5a+4b)(5b+4c)(5c+4a)}.$

### Solution 2

We have $(x-1)^2(25x^2+2x+4)\ge 0,\,$ $x\gt 0,\,$ in particular. This is equivalent to

$(5x^2+4)(5x^2+1)\ge 6x(8x^2+1).$

Letting $\displaystyle x=\sqrt{\frac{a}{b}}\,$ translates into

$(5a+4b)(5a+b)\ge 6\sqrt{ab}(8a+b).$

This shows that for any number of variables,

$\displaystyle \prod_{cycl}(5a+4b)(5a+b)\ge 6^n\prod_{cycl}a\prod_{cycl}(8a+b),$

where $n\,$ is the number of variables.

### Acknowledgment

Dan Sitaru has kindly shared a problem from the Romanian Mathematical Magazine, with a solution of his mailed on a LaTeX file, which I appreciate greatly. Solution 2 is by Leo Giugiuc.