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Pairwise Products

Problem

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 less than or equal 1/3.


Solution

Let x = 1/3 + a, y = 1/3 + b, and z = 1/3 + c.

therefore x + y + z = 1/3 + a + 1/3 + b + 1/3 + c = 1 + a + b + c.

But as x + y + z = 1, we deduce that a + b + c = 0.

therefore (a + b + c)2 = a2 + b2 + c2 + 2(ab + ac + bc) = 0
   2(ab + ac + bc) = -(a2 + b2 + c2)
therefore ab + ac + bc = -(a2 + b2 + c2)/2 = -d, where d greater than or equal 0

So 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 a + b + c = 0 and ab + ac + bc = -d, we get,
S = xy + xz + yz = 1/3 - d less than or equal 1/3   Q.E.D..

For a set of real numbers, R = {x1, x2, ... , xn} where x1 + x2 + ... +xn = 1, prove that the sum of pairwise products, S less than or equal (nminus1)/(2n).

Problem ID: 224 (24 May 2005)     Difficulty: 4 Star

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