Pairwise products

For any set of real numbers, R=x,y,zR = {x, y, z}, let sum of pairwise products, S=xy+xz+yzS = xy + xz + yz Given that x+y+z=1x + y + z = 1, prove that S≤13S\le \frac { 1 }{ 3 }

Note by Abdulrahman El Shafei
6 years, 5 months ago

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Comments

Letx=1/3+a,y=1/3+b,andz=1/3+c.x+y+z=1/3+a+1/3+b+1/3+c=1+a+b+c.Butasx+y+z=1,wededucethata+b+c=0.∴(a+b+c)2=a2+b2+c2+2(ab+ac+bc)=02(ab+ac+bc)=−(a2+b2+c2)∴ab+ac+bc=−(a2+b2+c2)/2=−d,whered≥0Soxy+xz+yz=(1/3+a)(1/3+b)+(1/3+a)(1/3+c)+(1/3+b)(1/3+c)=1/9+a/3+b/3+ab+1/9+a/3+c/3+ac+1/9+b/3+c/3+bc=1/3+(2/3)(a+b+c)+ab+ac+bcAsa+b+c=0andab+ac+bc=−d,weget,S=xy+xz+yz=1/3−d1/3=1/3+(2/3)(a+b+c)+ab+ac+bcAsa+b+c=0andab+ac+bc=−d,weget,S=xy+xz+yz=1/3−d≤1/3Let\quad x=1/3+a,y=1/3+b,and\quad z=1/3+c.\\ x+y+z=1/3+a+1/3+b+1/3+c=1+a+b+c.\\ But\quad as\quad x+y+z=1,we\quad deduce\quad that\quad a+b+c=0.\\ \therefore { (a+b+c) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+2(ab+ac+bc)=0\\ \quad \quad 2(ab+ac+bc)=-({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 })\\ \therefore \quad ab+ac+bc=-({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 })/2=-d,where\quad d\ge 0\\ So\quad xy+xz+yz\\ =(1/3+a)(1/3+b)+(1/3+a)(1/3+c)+(1/3+b)(1/3+c)\\ =1/9+a/3+b/3+ab+1/9+a/3+c/3+ac+1/9+b/3+c/3+bc\\ =1/3+(2/3)(a+b+c)+ab+ac+bc\\ As\quad a+b+c=0\quad and\quad ab+ac+bc=-d,we\quad get,\\ S=xy+xz+yz=1/3-d1/3=1/3+(2/3)(a+b+c)+ab+ac+bc\\ As\quad a+b+c=0\quad and\quad ab+ac+bc=-d,we\quad get,\\ S=xy+xz+yz=1/3-d\le 1/3

Abdulrahman El Shafei - 6 years, 5 months ago
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