# A Long Cyclic Inequality of Degree 4

### Lemma

If $a,b,c$ are the side lengths of a triangle with $a+b+c=6$ and $ab+bc+ca=3(4-t)$ then $0\le t\lt 1.$

Since $2a,2b,2c\lt a+b+c,$ $0\lt a,b,c\lt 3.$ It follows that $(3-a)(3-b)(3-c)\gt 0,$ implying

$27-9(a+b+c)+3(ab+bc+ca)\gt abc\gt 0,$

so that $-27+9(4-t)\gt 0,$ or $4-t\gt 3$ and, subsequently, $t\lt 1.$

From $(a+b+c)^2\ge 3(ab+bc+ca)$ we also have $t\ge 0.$

### Solution 1

With the lemma, the problem reduces to proving

$72(1-t)\ge 18(4-t)(1-t)$

which is, due to $0\le t\lt 1,$ equivalent to $4\ge 4-t$ and that is obvious.

### Solution 2

The inequality can be rewritten as

$\displaystyle \left[3\sum_{cycl}ab-\left(\sum_{cycl}a\right)^2\right]\left[4\sum_{cycl}ab-\left(\sum_{cycl}a\right)^2\right]\le 0,$

which obviously follows from

\displaystyle\begin{align}&3\sum_{cycl}ab-\left(\sum_{cycl}a\right)^2\le 0\\ &4\sum_{cycl}ab-\left(\sum_{cycl}a\right)^2\gt 0.\end{align}

To prove that, let $s$ be the semiperimeter of the triangle. Then

$(s-a)(s-b)(s-c)\gt 0,$

so that

\begin{align} s^3-(a+b+c)s^2+(ab+bc+ca)s-abc\gt 0. \end{align}

which is none other than,

$\displaystyle \left(\sum_{cycl}a\right)^3-2\left(\sum_{cycl}a\right)^3+4(ab+bc+ca)\sum_{cycl}a\gt 8abc\gt 0,$

so that $\displaystyle -\left(\sum_{cycl}a\right)^2+4(ab+bc+ca)\gt 0,$ and $\displaystyle 4(ab+bc+ca)\gt \left(\sum_{cycl}a\right)^2.$ On the other hand, $\displaystyle 3(ab+bc+ca)\le \left(\sum_{cycl}a\right)^2$ since it's equivalent to $\displaystyle \sum_{cycl}a^2\ge \sum_{cycl}ab.$

### Acknowledgment

The problem, with a solution (and Lemma), was kindly posted by Leo Giugiuc at the CutTheKnotMath facebook page. The problem is due to hoang giang.