# Leo Giugiuc's Cyclic Inequality in Three Variables

### Proof 1

Substitute $a-b=x,\,$ $b-c=y,\,$ $c-a=z.\,$ Then

$xy+yz+zx=ab+bc+ca-(a^2+b^2+c^2)$

and $x+y+z=0.\,$ We need toprove that

$\displaystyle 3\sqrt[3]{\frac{1}{4}x^2y^2z^2}+xy+yz+zx\le 0$

which is equivalent to $\displaystyle \sqrt[3]{\frac{1}{4}x^2y^2z^2}+xy-z^2\le 0.$

Use the AM-GM inequality for $\displaystyle \sqrt[3]{\frac{1}{4}x^2y^2z^2}\le \sqrt[3]{\frac{1}{4}\left(\frac{x+y}{4}\right)^2z^2}=\frac{z^2}{4}.$

Now, it is easy to prove $xy\le\displaystyle\frac{3}{4}z^2,\,$ which is equivalent to $0\le (x+y)^2+2(x^2+y^2)\,$ which is obviously correct.

### Proof 2

WLOG, $a\ge b\ge c.\,$ Let $a-b=x\ge 0,\,$ $b-c=y\ge 0,\,$ $a-c=x+y.\,$ We have

\displaystyle\begin{align} a^2+b^2+c^2-ab-bc-ca &=\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}\\ &=\frac{x^2+y^2+(x+y)^2}{2}=x^2+xy+y^2. \end{align}

The required inequality isequivalent to

$\displaystyle x^2+xy+y^2\ge 3\sqrt[3]{\frac{1}{4}x^2y^2(x+y)^2}.$

Now, using

$\displaystyle x^2+xy+y^2\ge\frac{3}{4}(x+y)^2\ge 3\sqrt[3]{\frac{1}{4}x^2y^2(x+y)^2},$

this reduces to $\displaystyle\left[\frac{(x+y)^2}{4}\right]^3\ge\frac{1}{4}x^2y^2(x+y)^2,\,$ or,

$\displaystyle (x+y)^2\left[(x+y)^4-4^2x^2y^2\right]\ge 0,$

and, subsequently, to

$\displaystyle (x+y)^2\left[(x+y)^2+4xy\right]\left[(x+y)^2-4xy\right]\ge 0,$

which is true.

### Proof 3

The required inequality is equivalent to

$\displaystyle\frac{1}{2}\left((a-b)^2+(b-c)^2+(c-a)^2\right)\ge 3\sqrt[3]{\frac{1}{4}(a-v)^2(b-c)^2(c-a)^2}.$

Introduce

$\displaystyle f(a,b,c)=\frac{1}{8}\left((a-b)^2+(b-c)^2+(c-a)^2\right)-\frac{27}{4}(a-b)^2(b-c)^2(c-a)^2.$

The previous inequality is equivalent to $f(a,b,c)\ge 0\,$ and, since $f(a,b,c)=f(0,b-a,c-a),\,$ suffice it to prove that $f(0,x,y)\ge 0.\,$ This is equivalent to

$4(x^2+y^2-xy)^3-27x^2y^2(x-y)^2=\ldots=(2x-y)^2(2y-x)^2(x+y)^2\ge 0.$

Equality is achieved for $\displaystyle a=\frac{b+c}{2}\,$ and cyclic permutations.

### Proof 4

Denote $a-b=x,\,$ $b-c=y,\,$ $c-a=z.\,$ Then $x+y+z=0\,$ and we need to show that

$\displaystyle\frac{x^2+y^2+z^2}{2}\ge 3\sqrt[3]{\frac{1}{4}x^2y^2z^2}$

which is equivalent to

$\displaystyle -(xy+yz+zx)\ge 3\sqrt[3]{\frac{1}{4}x^2y^2z^2}.$

Let $xy+yz+zx=-3q^2\,$ and $xyz=p.\,$ The last inequality is equivalent to $4q^6\ge p^2\,$ whose validity follows from the Cardano-Tartaglia solution of a 3rd degree equation with all three roots real.

### Acknowledgment

Leo Giugiuc has kindly posted this inequality from the mathematical inequalities facebook group at the CutTheknotMath facebook page, with several solutions. The inequality is due to Leo Giugiuc. Proof 1 si by Trán Hoàng Nam; Proof 2 is by Marian Dinca; Proof 3 is by Daniel Dan; Proof 4 is by Leo Giugiuc.