An AM-GM inequality (arithmetic mean - geometric mean inequality) Sn is
(x1 + x2 + … xn)·(1/n) ≥ (x1 · x2 … xn)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: Sn implies S2n

S2n is

(x1 + … xn + xn+1 + … x2n)·(1/2n) ≥ (x1 · … xn · xn+1 · … x2n)1/2n

(x1 + … xn + xn+1 + … x2n)·(1/n)·½ ≥ (x1 · … xn · xn+1 · … x2n)(1/n)·½

{(x1 + … xn)·(1/n) + (xn+1 + … x2n)·(1/n)} ½ ≥ {(x1 · … xn)1/n · (xn+1 · … x2n)1/n}½

Let a be the AM of x1…xn.
Let b be the AM of xn+1…x2n.
Let c be the GM of x1…xn.
Let d be the GM of xn+1…x2n.
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 S2n.

#### Induction 2: Sn implies Sn-1

Let A be the arithmetic mean and G be the geometric mean of the first n-1 terms, and fix xn = G. Then Sn can be written as:
(A(n-1) + G)/n ≥ (Gn-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 + … λn = 1 is:
Σ λixi ≥ Π xiλi     (for i = 1…n)
The left side is known as a convex combination.

#### Proof

Consider the function ƒ(x) = ex. ƒ is convex, therefore:
Σ λiƒ(ti) ≥ ƒ(Σ λiti)     (for i = 1…n)
This inequality, known as the Jensen's inequality, is sometimes used as the property that defines a convex function. Let ti = ln(xi) for positive xi's and substitute:
Σ λieln(xi) ≥ e (Σλiln(xi))     (for i = 1…n)
Σ λixi ≥ Π xi λi     (for i = 1…n)
QED