## 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 1/3.

#### Solution

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

`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.

(`a` + `b` + `c`)^{2} = `a`^{2} + `b`^{2} + `c`^{2} + 2(`ab` + `ac` + `bc`) = 0

2(`ab` + `ac` + `bc`) = -(`a`^{2} + `b`^{2} + `c`^{2})

`ab` + `ac` + `bc` = -(`a`^{2} + `b`^{2} + `c`^{2})/2 = -`d`, where `d` 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` 1/3 **Q.E.D.**.

For a set of real numbers, R = {`x`_{1}, `x`_{2}, ... , `x`_{n}} where `x`_{1} + `x`_{2} + ... +`x`_{n} = 1, prove that the sum of pairwise products, S (`n`1)/(2`n`).

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