# A Cyclic Inequality in Three Variables XIX

### Solution 1

Since the inequality is homogeneous we may assume $x+y+z=3.\,$ It then becomes:

$\displaystyle \frac{x}{(3-x)^3}+\frac{y}{(3-y)^3}+\frac{z}{(3-z)^3}\ge\frac{3}{8},$

$0\le x,y,z\le 3.\,$ Verify that

$\displaystyle\frac{x}{(3-x)^3}-\left(\frac{15x}{16}-\frac{3}{16}\right)=\frac{(x-1)^2(5x^2-38x+81)}{16(3-x)^3}\ge 0.$

Thus

\displaystyle \begin{align} \sum_{cycl}\frac{x}{(3-x)^3}&\ge\sum_{cycl}\frac{5x}{16}-\sum_{cycl}\frac{3}{16}\\ &=\frac{5(x+y+z)}{16}-\frac{9}{16}=\frac{6}{16}=\frac{3}{8}. \end{align}

Equality only when $x=y=z=1.$

### Solution 2

\displaystyle \begin{align} \sum_{cycl}\frac{x}{(y+z)^3} &= \sum_{cycl}\frac{x^4}{[x(y+z)]^3}\\ &\ge \frac{(x+y+z)^4}{[2(xy+yz+zx)]^3}. \end{align}

Therefore, suffice it to prove that $\displaystyle\frac{(x+y+z)^4}{(xy+yz+zx)^3}\ge\frac{27}{(x+y+z)^2},\,$ which is equivalent to

$(x+y+z)^2\ge 3(xy+yz+zx),$

This, in turn, is equivalent to $x^2+y^2+z^2\ge xy+yz+zx\;$ and is proved by, say, rearranegment.

### Acknowledgment

This is problem SP053 from the Spring 2017 issue of the Romanian Mathematical Magazine, posted by D. M. Batinetu (Romania.) Solution 1 is by Imad Zak (Lebanon).