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,
with equality if and only if . This follows from the fact that the square of a real number is always non-negative (greater than or equal to zero) and from the identity :
Hence , with equality when , i.e. . The AM–GM inequality then follows from taking the positive square root of both sides and then dividing both sides by 2.
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 V, Proposition 25.
Extensions of the AM–GM inequality treat weighted means and generalized means.
Background
The arithmetic mean, or less precisely the average, of a list of numbers is the sum of the numbers divided by :
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:
If , this is equal to the exponential of the arithmetic mean of the natural logarithms of the numbers:
The inequality
Restating the inequality using mathematical notation, we have that for any list of nonnegative real numbers ,
and that equality holds if and only if .
Geometric interpretation
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, is the edge length of an -dimensional cube of equal volume, which therefore is also the average length of edges incident to a vertex of the cube.
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.
Examples
Example 1
If , then the AM-GM inequality tells us that
Example 2
A simple upper bound for can be found. AM-GM tells us
and so
with equality at .
Equivalently,
Example 3
Consider the function
for all positive real numbers , and . Suppose we wish to find the minimal value of this function. It can be rewritten as:
with
Applying the AM–GM inequality for , we get
Further, we know that the two sides are equal exactly when all the terms of the mean are equal:
All the points satisfying these conditions lie on a half-line starting at the origin and are given by
Applications
Cauchy-Schwarz inequality The AM-GM equality can be used to prove the Cauchy–Schwarz inequality.
Annualized returns 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.
Graph theory In graph theory, the AM-GM inequality can be used to prove that the number of edges of a bipartite graph on vertices is bounded from above by . The vertices of a bipartite graph with vertices may be divided into two classes and . Since we can only draw edges between vertices in one class to another, the maximum number of edges is given by . By the AM-GM inequality, we can bound from above like so:
Nonnegative polynomials 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 , , and , that is, Simplifying and multiplying both sides by 3 gives soAaron Potechin, Sum of Squares seminar, University of Chicago, "Lecture 5: SOS Proofs and the Motzkin Polynomial", slide 25
Proofs of the AM–GM inequality The AM–GM inequality can be proven in many ways.
Proof using Jensen's inequality
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
Taking antilogs (the exponential) of the far left and far right sides, we have the AM–GM inequality.
Proof by successive replacement of elements
We have to show that
with equality only when all numbers are equal.
If not all numbers are equal, then there exist such that . Replacing by and by will leave the arithmetic mean of the numbers unchanged, but will increase the geometric mean because
If the numbers are still not equal, we continue replacing numbers as above. After at most such replacement steps all the numbers will have been replaced with while the geometric mean strictly increases at each step. After the last step, the geometric mean will be , proving the inequality.
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 such that . Replacing by and by leaves the geometric mean unchanged but strictly decreases the arithmetic mean since
. The proof then follows along similar lines as in the earlier replacement.
Induction proofs
Proof by induction #1
Of the non-negative real numbers , the AM–GM statement is equivalent to
with equality if and only if for all .
For the following proof we apply mathematical induction and only well-known rules of arithmetic.
Induction basis: For the statement is true with equality.
Induction hypothesis: Suppose that the AM–GM statement holds for all choices of non-negative real numbers.
Induction step: Consider non-negative real numbers , . Their arithmetic mean satisfies
If all the are equal to , then we have equality in the AM–GM statement and we are done. In the case where some are not equal to , there must exist one number that is greater than the arithmetic mean , and one that is smaller than . Without loss of generality, we can reorder our in order to place these two particular elements at the end: and . Then
Now define with
and consider the numbers which are all non-negative. Since
Thus, is also the arithmetic mean of numbers and the induction hypothesis implies
Due to (*) we know that
hence
in particular . Therefore, if at least one of the numbers is zero, then we already have strict inequality in (**). Otherwise the right-hand side of (**) is positive and strict inequality is obtained by using the estimate (***) to get a lower bound of the right-hand side of (**). Thus, in both cases we can substitute (***) into (**) to get
which completes the proof.
Proof by induction #2
First of all we shall prove that for real numbers and there follows
Indeed, multiplying both sides of the inequality by , gives
whence the required inequality is obtained immediately.
Now, we are going to prove that for positive real numbers satisfying , there holds
The equality holds only if .
Induction basis: For the statement is true because of the above property.
Induction hypothesis: Suppose that the statement is true for all natural numbers up to .
Induction step: Consider natural number , i.e. for positive real numbers , there holds . There exists at least one , so there must be at least one . Without loss of generality, we let and .
Further, the equality we shall write in the form of . Then, the induction hypothesis implies
However, taking into account the induction basis, we have
which completes the proof.
For positive real numbers , let's denote
The numbers satisfy the condition . So we have
whence we obtain
with the equality holding only for .
Proof by Cauchy using forward–backward induction
The following proof by cases relies directly on well-known rules of arithmetic but employs the rarely used technique of forward-backward-induction. It is essentially from Augustin Louis Cauchy and can be found in his Cours d'analyse.
The case where all the terms are equal
If all the terms are equal:
then their sum is , so their arithmetic mean is ; and their product is , so their geometric mean is ; therefore, the arithmetic mean and geometric mean are equal, as desired.
The case where not all the terms are equal
It remains to show that if not all the terms are equal, then the arithmetic mean is greater than the geometric mean. Clearly, this is only possible when .
This case is significantly more complex, and we divide it into subcases.
The subcase where n = 2
If , then we have two terms, and , and since (by our assumption) not all terms are equal, we have:
hence
as desired.
The subcase where n = 2k
Consider the case where , where is a positive integer. We proceed by mathematical induction.
In the base case, , so . We have already shown that the inequality holds when , so we are done.
Now, suppose that for a given , we have already shown that the inequality holds for , and we wish to show that it holds for . To do so, we apply the inequality twice for numbers and once for numbers to obtain:
where in the first inequality, the two sides are equal only if
and
(in which case the first arithmetic mean and first geometric mean are both equal to , and similarly with the second arithmetic mean and second geometric mean); and in the second inequality, the two sides are only equal if the two geometric means are equal. Since not all numbers are equal, it is not possible for both inequalities to be equalities, so we know that:
as desired.
The subcase where n < 2k
If is not a natural power of , then it is certainly less than some natural power of 2, since the sequence is unbounded above. Therefore, without loss of generality, let be some natural power of that is greater than .
So, if we have terms, then let us denote their arithmetic mean by , and expand our list of terms thus:
We then have:
so
and
as desired.
Proof by induction using basic calculus
The following proof uses mathematical induction and some basic differential calculus.
Induction basis: For the statement is true with equality.
Induction hypothesis: Suppose that the AM–GM statement holds for all choices of non-negative real numbers.
Induction step: In order to prove the statement for non-negative real numbers , we need to prove that
with equality only if all the numbers are equal.
If all numbers are zero, the inequality holds with equality. If some but not all numbers are zero, we have strict inequality. Therefore, we may assume in the following, that all numbers are positive.
We consider the last number as a variable and define the function
Proving the induction step is equivalent to showing that for all , with only if and are all equal. This can be done by analyzing the critical points of using some basic calculus.
The first derivative of is given by
A critical point has to satisfy , which means
After a small rearrangement we get
and finally
which is the geometric mean of . This is the only critical point of . Since for all , the function is strictly convex and has a strict global minimum at . Next we compute the value of the function at this global minimum: