In mathematics, the inequality of arithmetic and geometric means, or more briefly the AM–GM inequality, states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list; and further, that the two means are equal if and only if every number in the list is the same (in which case they are both that number).
The simplest non-trivial case is for two non-negative numbers and , that is,
x+y | |
2 |
\ge\sqrt{xy}
\begin{align} 0&\le(x-y)2\\ &=x2-2xy+y2\\ &=x2+2xy+y2-4xy\\ &=(x+y)2-4xy. \end{align}
For a geometrical interpretation, consider a rectangle with sides of length and ; it has perimeter and area . Similarly, a square with all sides of length has the perimeter and the same area as the rectangle. The simplest non-trivial case of the AM–GM inequality implies for the perimeters that and that only the square has the smallest perimeter amongst all rectangles of equal area.
The simplest case is implicit in Euclid's Elements, Book 5, Proposition 25.[1]
Extensions of the AM–GM inequality treat weighted means and generalized means.
The arithmetic mean, or less precisely the average, of a list of numbers is the sum of the numbers divided by :
x1+x2+ … +xn | |
n |
.
The geometric mean is similar, except that it is only defined for a list of nonnegative real numbers, and uses multiplication and a root in place of addition and division:
\sqrt[n]{x1 ⋅ x2 … xn}.
If, this is equal to the exponential of the arithmetic mean of the natural logarithms of the numbers:
\exp\left(
ln{x1 | |
+ |
ln{x2}+ … +ln{xn}}{n}\right).
Restating the inequality using mathematical notation, we have that for any list of nonnegative real numbers,
x1+x2+ … +xn | |
n |
\ge\sqrt[n]{x1 ⋅ x2 … xn},
and that equality holds if and only if .
In two dimensions, is the perimeter of a rectangle with sides of length and . Similarly, is the perimeter of a square with the same area,, as that rectangle. Thus for the AM–GM inequality states that a rectangle of a given area has the smallest perimeter if that rectangle is also a square.
The full inequality is an extension of this idea to dimensions. Consider an -dimensional box with edge lengths . Every vertex of the box is connected to edges of different directions, so the average length of edges incident to the vertex is .On the other hand,
\sqrt[n]{x1x2 … xn}
Thus the AM–GM inequality states that only the -cube has the smallest average length of edges connected to each vertex amongst all -dimensional boxes with the same volume.[2]
If
a,b,c>0
(1+a)(1+b)(1+c)\ge2\sqrt{1 ⋅ {a}} ⋅ 2\sqrt{1 ⋅ {b}} ⋅ 2\sqrt{1 ⋅ {c}}=8\sqrt{abc}
A simple upper bound for
n!
1+2+...+n\gen\sqrt[n]{n!}
n(n+1) | |
2 |
\gen\sqrt[n]{n!}
and so
\left( | n+1 |
2 |
\right)n\gen!
with equality at
n=1
Equivalently,
(n+1)n\ge2nn!
Consider the function
f(x,y,z)=
x | |
y |
+\sqrt{
y | |
z |
for all positive real numbers, and . Suppose we wish to find the minimal value of this function. It can be rewritten as:
\begin{align} f(x,y,z) &=6 ⋅
| ||||||||||||
+ |
1 | \sqrt{ | |
2 |
y | |
z |
x | ||||
|
, x2=x
\sqrt{ | |||||
|
y | |
z |
Applying the AM–GM inequality for, we get
\begin{align} f(x,y,z) &\ge6 ⋅ \sqrt[6]{
x | |
y |
⋅
1 | \sqrt{ | |
2 |
y | |
z |
Further, we know that the two sides are equal exactly when all the terms of the mean are equal:
f(x,y,z)=22/3 ⋅ 31/2 when
x | |
y |
=
1 | \sqrt{ | |
2 |
y | |
z |
All the points satisfying these conditions lie on a half-line starting at the origin and are given by
(x,y,z)=r(t,\sqrt[3]{2}\sqrt{3}t, | 3\sqrt{3 |
The AM-GM equality can be used to prove the Cauchy–Schwarz inequality.
In financial mathematics, the AM-GM inequality shows that the annualized return, the geometric mean, is less than the average annual return, the arithmetic mean.
The Motzkin polynomial is a nonnegative polynomial which is not a sum of square polynomials. It can be proven nonnegative using the AM-GM inequality with
x1=x4y2
x2=x2y4
x3=1
The AM–GM inequality can be proven in many ways.
Jensen's inequality states that the value of a concave function of an arithmetic mean is greater than or equal to the arithmetic mean of the function's values. Since the logarithm function is concave, we have
log\left(
\sumxi | |
n |
\right)\geq
1 | |
n |
\sumlogxi=
1 | |
n |
log\left(\prodxi\right)=log\left(\prod
1/n | |
x | |
i\right) |
.
Taking antilogs of the far left and far right sides, we have the AM–GM inequality.
We have to show that
\alpha=
x1+x2+ … +xn | |
n |
\ge\sqrt[n]{x1x2 … xn}=\beta
with equality only when all numbers are equal.
If not all numbers are equal, then there exist
xi,xj
xi<\alpha<xj
\alpha
(xi+xj-\alpha)
\alpha(xj+xi-\alpha)-xixj=(\alpha-xi)(xj-\alpha)>0
If the numbers are still not equal, we continue replacing numbers as above. After at most
(n-1)
\alpha
\sqrt[n]{\alpha\alpha … \alpha}=\alpha
It may be noted that the replacement strategy works just as well from the right hand side. If any of the numbers is 0 then so will the geometric mean thus proving the inequality trivially. Therefore we may suppose that all the numbers are positive. If they are not all equal, then there exist
xi,xj
0<xi<\beta<xj
xi
\beta
xj
xixj | |
\beta |
xi+xj-\beta-
xixj | |
\beta |
=
(\beta-xi)(xj-\beta) | |
\beta |
>0
Of the non-negative real numbers, the AM–GM statement is equivalent to
\alphan\gex1x2 … xn