A Cyclic Inequality on [-1,1]

Problem

Solution 1

Case 1: $-1\in\{x,y,z\},$ say $z=-1.$ The inequality then reduces to $(1-x)(1-y)\ge 0,$ which is obvious. Equality is attained at $(1,k,-1)$ or $(k,1,-1),$ for any $|k|\le 1.$

Case 2: none of $x,y,z$ equals $-1.$ Let $\displaystyle \frac{1-x}{1+x}=a,$ $\displaystyle \frac{1-y}{1+y}=b,$ $\displaystyle \frac{1-z}{1+z}=c.$ Then $a,b,c\ge 0$ and $\displaystyle x=\frac{1-a}{1+a},$ $\displaystyle y=\frac{1-b}{1+b},$ $\displaystyle z=\frac{1-c}{1+c}.$ The required inequality reduces to $4+4abc\ge 0,$ which is obvious.

Solution 2

$(x+1)(y+1)(z+1)-(x-1)(y-1)(z-1) \ge 0.$ Therefore, $2xy+2yz+2zx +2 \ge 0.$

Solution 3

Add $y^2$ to both sides of the desired inequality:

$y^2+xy+yz+zx=(y+x)(y+z)\ge (y+1)(y-1)=y^2-1.$

On the left, a parabola in $y$ which is negative on, say, $[-x,-z]\subset [-1,1]$ which is where the parabola on the right is negative. The two parabolas are of the same shape. Thus the inequality indeed holds.

Acknowledgment

I am grateful to Leo Giugiuc for mailing me this problem by Nguyen Viet Hung. The problem has been originally posted at the Olimpiada pe Scoala (The School Yard Olympiad).

Solution 1 is by Leo Giugiuc; Solution 2 is by Nguyễn Đang Quà; Solution 3 is by Long Huynh Huu.

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