A Simple Cyclic Inequality in Three Variables

Preliminaries

It is hard to overlook the fact that

$\displaystyle \sum_{cycl}(a^2+b^2-c^2)^2=3\sum_{cycl}a^4+2\sum_{cycl}a^2b^2,$

while

$\displaystyle 3\left(\sum_{cycl}a^2\right)^2=3\sum_{cycl}a^4+6\sum_{cycl}a^2b^2,$

which reduces the required inequality to proving

$\displaystyle \sum_{cycl}a^2b^2\ge abc\sum_{cycl}a.$

The submitted proofs have differed in establishing that inequality.

Solution 1

$(xy-yz)^2\ge 0\,\Rightarrow\,x^2y^2+y^2z^2\ge 2xy^2z.\,$ Thus we have

\begin{align} &a^2b^2+b^2c^2\ge 2ab^2c\\ &b^2c^2+c^2a^2\ge 2abc^2\\ &c^2a^2+a^2b^2\ge 2a^2bc. \end{align}

It follows that $\displaystyle 2\sum_{cycl}a^2b^2\ge 2abc\sum_{cycl}a.\,$

Solution 2

The inequality $x^2+y^2+z^2\ge xy+yz+zx\,$ is well known (e.g., by the Rearrangement inequality.) Substitute $x=ab,\,$ $y=bc,\,$ $z=ca.\,$

Solution 3

Since the triple $(2,2,0)\,$ majorizes $(2,1,1),\,$ $\displaystyle \sum_{cycl}a^2b^2\ge abc\sum_{cycl}a$ is a particular case of Muirhead's inequality.

Acknowledgment

Dan Sitaru has kindly posted the problem from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page and later communicated several solutions. Solution 1 is by Abdul Aziz and, independently, by Seyran Ibrahimov; Solution 2 is by Kevin Soto Palacios and, independently, by Soumava Chakraborty. Kevin Soto Palacios has observed that the restriction to positive numbers is unnecessary.