# Pairwise products

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

Note by Abdulrahman El Shafei
3 years, 9 months ago

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$$Let\quad x=1/3+a,y=1/3+b,and\quad z=1/3+c.\\ x+y+z=1/3+a+1/3+b+1/3+c=1+a+b+c.\\ But\quad as\quad x+y+z=1,we\quad deduce\quad that\quad a+b+c=0.\\ \therefore { (a+b+c) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+2(ab+ac+bc)=0\\ \quad \quad 2(ab+ac+bc)=-({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 })\\ \therefore \quad ab+ac+bc=-({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 })/2=-d,where\quad d\ge 0\\ So\quad xy+xz+yz\\ =(1/3+a)(1/3+b)+(1/3+a)(1/3+c)+(1/3+b)(1/3+c)\\ =1/9+a/3+b/3+ab+1/9+a/3+c/3+ac+1/9+b/3+c/3+bc\\ =1/3+(2/3)(a+b+c)+ab+ac+bc\\ As\quad a+b+c=0\quad and\quad ab+ac+bc=-d,we\quad get,\\ S=xy+xz+yz=1/3-d1/3=1/3+(2/3)(a+b+c)+ab+ac+bc\\ As\quad a+b+c=0\quad and\quad ab+ac+bc=-d,we\quad get,\\ S=xy+xz+yz=1/3-d\le 1/3$$

- 3 years, 8 months ago