# Dorin Marghidanu's Cyclic Inequality in Three Variables II

### Solution 1

Denote $a+b+c=p,\,$ $ab+bc+ca=q,\,$ $abc=r.\,$ By the AM-GM inequality, $p\ge 3r^{\frac{1}{3}}\,$ while $q\ge 3r^{\frac{2}{3}},\,$ such that

(1)

$pq\ge 9r.$

We also know (or can easily verify) that

(2)

$\displaystyle \sum_{cycl}ab(a+b)=pq-3r$

and

(3)

$\displaystyle \prod_{cycl}(a+b)=pq-r.$

Now,

\displaystyle \begin{align} LHS &= \frac{\displaystyle \sum_{cycl}ab(a+b)}{\displaystyle \prod_{cycl}(a+b)}\\ &=\frac{pq-3r}{pq-r}. \end{align}

Thus, the required inequality is equivalent to $\displaystyle \frac{pq-3r}{pq-r}\ge\frac{3}{4},\,$ or

$4pq-12r\ge 3pq-3r,$

i.e., $pq\ge 9r,\,$ which is true according to (1). Equality is attained at $a=b=c.$

### Solution 2

Using substitutions $a=\displaystyle \frac{1}{x},\,$ $b=\displaystyle \frac{1}{y},\,$ $c=\displaystyle \frac{1}{z},\,$ and Bergstrom's inequality, we have

\displaystyle \begin{align} LHS &= \sum_{cycl}\frac{z^2}{x+z)(y+z)}\\ &\ge\frac{(x+y+z)^2}{x^2+y^2+z^2+3(xy+yz+zx)}\\ &=\frac{(x+y+z)^2}{(x+y+z)^2+(xy+yz+zx)}\\ &\ge\frac{(x+y+z)^2}{(x+y+z)^2+\frac{1}{3}(x+y+z)^2}\\ &=\frac{3}{4}. \end{align}

### Solution 3

Multiply out:

$4[ab(a+b)+bc(b+c)+ca(c+a)]\ge 3(a+b)(b+c)(c+a).$

Do some algebra to reduce this to

$(a^2b+b^2c+c^2a)+(ab^2+bc^2+ca^2)-6abc\ge 0.$

But this is true because of the AM-GM inequality:

$a^2b+b^2c+c^2a\ge 3abc\,$ and $ab^2+bc^2+ca^2\ge 3abc.$

### Acknowledgment

This problem has been kindly posted at the CutTheKnotMath facebook page by Dorin Marghidanu. Solution is by Imad Zak; Solution 2 is by Dorin Marghidanu; Solution 3 is by Srinivas Vemuri.