An
AM-GM inequality
(
arithmetic mean -
geometric mean inequality)
S
n 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 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: Sn implies S2n
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: Sn implies Sn-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 ≥ (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 λ
1 + … λ
n = 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) = e
x. 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 t
i = ln(x
i) for positive x
i's and substitute:
Σ λieln(xi) ≥
e (Σλiln(xi))
(for i = 1…n)
Σ λixi ≥
Π xi λi
(for i = 1…n)
QED