## Mean Proof

#### Problem

Prove that (`a` + `b`)/2 (`ab`), where `a`, `b` are non-negative real numbers.

#### Solution

Without loss of generality, assume that `a` `b`; let `b` = `a` + `k`, where `k` 0.

Therefore, (`a` + `b`)/2 = (`a` + `a` + `k`)/2 = (2`a` + `k`)/2 = `a` + `k`/2.

Also, `ab` = `a`(`a` + `k`) = `a`^{2} + `ak` = (`a` + `k`/2)^{2} `k`^{2}/4.

Clearly (`a` + `k`/2)^{2} (`a` + `k`/2)^{2} `k`^{2}/4 = `ab`; that is, ((`a` + `b`)/2)^{2} `ab`.

Hence (`a` + `b`)/2 (`ab`).

Alternatively we begin with the observation that (`a` `b`)^{2} 0.

Expanding we get, `a` + `b` 2(`ab`) 0, which leads to (`a` + `b`)/2 (`ab`).

Prove that (`a` + `b` + `c`)/3 ^{3}(`abc`).

Can you generalise?

Problem ID: 230 (10 Jul 2005) Difficulty: 3 Star