Leo Giugiuc's Cyclic Inequality in Three Variables


Leo Giugiuc's Cyclic Inequality in Three Variables

Proof 1

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


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

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


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

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.


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.


Cyclic inequalities in three variables

|Contact| |Front page| |Contents| |Algebra| |Store|

Copyright © 1996-2017 Alexander Bogomolny


Search by google: