Tran Hoang Nam's Cyclic Inequality in Three Variables

Proof 1

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

$(x^2+y^2+z^2)^3+8(x^3y^3+y^3z^3+z^3x^3)\ge 0.$

Since $x+y+z=0,\,$ $x^2+y^2+z^2=-2(xy+yz+zx)\,$ and $x^3y^3+y^3z^3+z^3x^3=(xy+yz+zx)^3-3x^2y^2z^2.$

Hence, the inequality is equivalent to

$-8(x^3y^3+y^3z^3+z^3x^3)+8(x^3y^3+y^3z^3+z^3x^3)+24x^2y^2z^2\ge 0.$

Obviously true. Equality is achieved for $(a-b)(b-c)(c-a)=0.$

Proof 2

Define $\displaystyle f(a,b,c)=\frac{1}{8}\left(\sum_{cycl}(a-b)^2\right)^3-\sum_{cycl}(a-b)^3(a-c)^3.\,$ We need to show that $f(a,b,c)\ge 0.$

WLOG, $a\le b\le c,\,$ $f(a,b,c)=f(a-a,b-a,c-a)=f(0,x,y),\,$ $x,y\ge 0.$

\begin{align} f(0,x,y) &+ (x^2+y^2-xy)^3-x^3y^3-x^3(x-y)^3-y^3(y-x)^3\\ &=3x^2y^2(x-y)^2\\ &\ge 0. \end{align}

Which is true.

Proof 3

WLOG, suppose $a\le b\le c\,$ and let $b-a=x\gt 0\,$ and $c-b=y\gt 0,\,$ implying $c-a=x+y\,$ and we remark that $\displaystyle\sum_{cycl}a^2-\sum_{cycl}ab=\frac{1}{2}\sum_{cycl}(a-b)^2.\,$ The $\displaystyle RHS=\frac{1}{8}(x^2+y^2+(x+y)^2)^3\,$ while the $LHS=x^3(x+y)^3-x^3y^3+y^3(x+y)^3.\,$ Now $RHS-LHS=24x^2y^2(x+y)^2\ge 0.\,$ Q.E.D.

Equality when $x=0\,$ or $y=0\,$ or $x=y=0,\,$ i.e., $a=b\,$ or $b=c\,$ or $a=b=c.$

Acknowledgment

Leo Giugiuc has kindly posted at the CutTheknotMath facebook page this inequality from the mathematical inequalities facebook group. The inequality is due to Trán Hoàng Nam. Proof 1 is by Leo Giugiuc; Proof 2 is by Daniel Dan; Proof 3 is by Imad Zak.