An

AM-GM inequality
(

arithmetic mean -

geometric mean inequality)
S

_{n} is

(x_{1} + x_{2} + … x_{n})·(1/n) ≥
(x_{1} · x_{2} … 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 S

_{2}.
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 S

_{1} is an equality.

#### Induction 1: S_{n} implies S_{2n}

S

_{2n} is

(x

_{1} + … x

_{n} + x

_{n+1} + … x

_{2n})·(1/2n)
≥
(x

_{1} · … x

_{n} · x

_{n+1} · … x

_{2n})

^{1/2n}
(x

_{1} + … x

_{n} + x

_{n+1} + … x

_{2n})·(1/n)·½
≥
(x

_{1} · … x

_{n} · x

_{n+1} · … x

_{2n})

^{(1/n)·½}
{(x

_{1} + … x

_{n})·(1/n) +
(x

_{n+1} + … x

_{2n})·(1/n)} ½
≥
{(x

_{1} · … x

_{n})

^{1/n}
· (x

_{n+1} · … x

_{2n})

^{1/n}}

^{½}
Let a be the AM of x

_{1}…x

_{n}.

Let b be the AM of x

_{n+1}…x

_{2n}.

Let c be the GM of x

_{1}…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 S

_{2},

(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 λ

_{1} + … λ

_{n} = 1
is:

Σ λ_{i}x_{i} ≥ Π x_{i}^{λi} (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_{i}ƒ(t_{i}) ≥
ƒ(Σ λ_{i}t_{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_{i}e^{ln(xi)} ≥
e ^{(Σλiln(xi))}
(for i = 1…n)

Σ λ_{i}x_{i} ≥
Π x_{i} ^{λi}
(for i = 1…n)

QED