proof of the AM-GM inequality (thing)

An AM-GM inequality (arithmetic mean - geometric mean inequality) S_n is

(x₁ + x₂ + … x_n)·(1/n) ≥ (x₁ · x₂ … x_n)^1/n

where x _i nonnegative for all i.

The left side is the arithmetic mean and the right side is the geometric mean. {x _i} cannot be extended to the entire reals. A counterexample to such would be (-1, -1, 2). If one x _i is zero, then the GM is zero, in which case the inequality holds since the AM is nonnegative. The induction proof will cover only the case where each x _i is nonnegative and non-zero.

Base case

Let n = 2.

(a - b)² ≥ 0
a² - 2ab + b² ≥ 0
a² + 2ab + b² ≥ 4ab
(a + b)² ≥ 4ab
(a + b)/2 ≥ (ab)^1/2

The last inequality is S₂. Another way to show this is a right triangle with lengths (a - b)/2, (ab)^1/2, and (a+b)/2 as the hypotenuse. The case for S₁ is an equality.

Induction 1: S_n implies S_2n

S_2n is

(x₁ + … x_n + x_n+1 + … x_2n)·(1/2n) ≥ (x₁ · … x_n · x_n+1 · … x_2n)^1/2n

(x₁ + … x_n + x_n+1 + … x_2n)·(1/n)·½ ≥ (x₁ · … x_n · x_n+1 · … x_2n)^(1/n)·½

{(x₁ + … x_n)·(1/n) + (x_n+1 + … x_2n)·(1/n)} ½ ≥ {(x₁ · … x_n)^1/n · (x_n+1 · … x_2n)^1/n}^½

Let a be the AM of x₁…x_n.
Let b be the AM of x_n+1…x_2n.
Let c be the GM of x₁…x_n.
Let d be the GM of x_n+1…x_2n.
Then the above inequality can be rewritten as:

(a + b)/2 ≥ (cd)^½

By the induction hypothesis, we know a ≥ c and b ≥ d. Hence (ab)^½ ≥ (cd)^½ since the square root function is monotone increasing. Then by S₂,

(a+b)/2 ≥ (ab)^½ ≥ (cd)^½

which, without the middle expression, is S_2n.

Induction 2: S_n implies S_n-1

Let A be the arithmetic mean and G be the geometric mean of the first n-1 terms, and fix x_n = G. Then S_n can be written as:

(A(n-1) + G)/n ≥ (G^n-1·G)^1/n
A (n-1) + G ≥ nG
A (n-1) ≥ G (n - 1)
A ≥ G

This completes the proof.

A Generalization

A generalized A-G mean inequality with nonnegative weights λ₁ + … λ_n = 1 is:

Σ λ_ix_i ≥ Π x_i^λ_i     (for i = 1…n)

The left side is known as a convex combination.

Proof

Consider the function ƒ(x) = e^x. ƒ is convex, therefore:

Σ λ_iƒ(t_i) ≥ ƒ(Σ λ_it_i)     (for i = 1…n)

This inequality, known as the Jensen's inequality, is sometimes used as the property that defines a convex function. Let t_i = ln(x_i) for positive x_i's and substitute:

Σ λ_ie^ln(x_i) ≥ e ^{(Σλ_iln(x_i))}     (for i = 1…n)
Σ λ_ix_i ≥ Π x_i ^λ_i     (for i = 1…n)

QED