Inequality of arithmetic and geometric means
Encyclopedia
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 number
s 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.
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 x1, x2, . . ., xn > 0, this is equal to the exponential
of the arithmetic mean of the natural logarithm
s of the numbers:
and that equality holds if and only if x1 = x2 = . . . = xn.
of a rectangle with sides of length x1 and x2. Similarly, 4√(x1x2) is the perimeter of a square with the same area
. Thus for n = 2 the AM-GM inequality states that a square has the smallest perimeter among all rectangles with equal area.
The full inequality is an extension of this idea to n dimensions. Every vertex on an n-dimensional box is connected to n edges. If these edges are given lengths x1, x2, ..., xn, then x1 + x2 + … + xn is the total length of edges connected to the vertex. There are 2n vertices, so multiply this by 2n; however, since each edge meets two vertices, that counts every edge twice. Thus we divide by 2 and conclude that there are n2n−1 edges. There are equally many edges of each length and n lengths; hence there are 2n−1 edges of each length. The total edge-length is therefore 2n−1(x1 + x2 + … + xn). On the other hand,
is the total length of edges connected to a vertex on an n-dimensional cube of equal volume. Since the inequality says
we get
Thus the AM–GM inequality states that an n-cube
has the smallest sum of lengths of edges connected to each vertex among all n-dimensional boxes with the same volume.
Other generalizations of the inequality of arithmetic and geometric means include these:
for x, y, and z all positive real numbers. Suppose we wish to find the minimum value of this function. Rewriting a bit, and applying the AM–GM inequality, we have:
Further, we know that the two sides are equal exactly when all the terms of the mean are equal:
, using the concave function ln(x). It can also be proven using the rearrangement inequality. Considering length and required prerequisites, the proof by induction given below is probably the best recommendation for first reading.
With the arithmetic mean
of the non-negative real numbers x1,...,xn, the AM–GM statement is equivalent to
with equality if and only if μ = xi for all i = 1,...,n.
For the following proof we apply mathematical induction
and only well-known rules of arithmetic.
Induction basis: For n = 1 the statement is true with equality.
Induction hypothesis: Suppose that the AM–GM statement holds for all choices of n non-negative real numbers.
Induction step: Consider n + 1 non-negative real numbers. Their arithmetic mean μ satisfies
If all numbers are equal to μ, then we have equality in the AM–GM statement and we are done. Otherwise we may find one number that is greater than μ and one that is smaller than μ, say xn > μ and xn+1 < μ. Then
Now consider the n numbers
with 
which are also non-negative. Since
μ is also the arithmetic mean of
and the induction hypothesis implies
Due to (*) we know that
hence
in particular μ > 0. Therefore, if at least one of the numbers x1,...,xn−1 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 get
which completes the proof.
provided a proof similar to what follows. Let f(x) = ex−1 − x, with derivative f(x) = ex−1 − 1. Observe f(1) = 0 and hence that f has an absolute minimum of f(1) = 0. Now x ≤ ex−1 for all real x.
Consider a list of non-negative numbers
with arithmetic mean μ. By repeated application of the above inequality, we obtain the following:
But the exponential argument can be simplified:
Returning to (1),
which produces the result:
and can be found in his Cours d'analyse.
then their sum is nx1, so their arithmetic mean is x1; and their product is x1n, so their geometric mean is x1; therefore, the arithmetic mean and geometric mean are equal, as desired.
This case is significantly more complex, and we divide it into subcases.
The subcase where n
If n = 2, then we have two terms, x1 and x2, and since (by our assumption) not all terms are equal, we have:
Mathematics
Mathematics is the study of quantity, space, structure, and change. Mathematicians seek out patterns and formulate new conjectures. Mathematicians resolve the truth or falsity of conjectures by mathematical proofs, which are arguments sufficient to convince other mathematicians of their validity...
, the inequality of arithmetic and geometric means, or more briefly the AM–GM inequality, states that the arithmetic mean
Arithmetic mean
In mathematics and statistics, the arithmetic mean, often referred to as simply the mean or average when the context is clear, is a method to derive the central tendency of a sample space...
of a list of non-negative real number
Real number
In mathematics, a real number is a value that represents a quantity along a continuum, such as -5 , 4/3 , 8.6 , √2 and π...
s is greater than or equal to the geometric mean
Geometric mean
The geometric mean, in mathematics, is a type of mean or average, which indicates the central tendency or typical value of a set of numbers. It is similar to the arithmetic mean, except that the numbers are multiplied and then the nth root of the resulting product is taken.For instance, the...
of the same list; and further, that the two means are equal if and only if
If and only if
In logic and related fields such as mathematics and philosophy, if and only if is a biconditional logical connective between statements....
every number in the list is the same.
Background
The arithmetic mean, or less precisely the average, of a list of n numbers x1, x2, . . ., xn is the sum of the numbers divided by 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:
If x1, x2, . . ., xn > 0, this is equal to the exponential
Exponential function
In mathematics, the exponential function is the function ex, where e is the number such that the function ex is its own derivative. The exponential function is used to model a relationship in which a constant change in the independent variable gives the same proportional change In mathematics,...
of the arithmetic mean of the natural logarithm
Natural logarithm
The natural logarithm is the logarithm to the base e, where e is an irrational and transcendental constant approximately equal to 2.718281828...
s of the numbers:
The inequality
Restating the inequality using mathematical notation, we have that for any list of n nonnegative real numbers x1, x2, . . ., xn,and that equality holds if and only if x1 = x2 = . . . = xn.
Geometric interpretation
In two dimensions, 2x1 + 2x2 is the perimeterPerimeter
A perimeter is a path that surrounds an area. The word comes from the Greek peri and meter . The term may be used either for the path or its length - it can be thought of as the length of the outline of a shape. The perimeter of a circular area is called circumference.- Practical uses :Calculating...
of a rectangle with sides of length x1 and x2. Similarly, 4√(x1x2) is the perimeter of a square with the same area
Area
Area is a quantity that expresses the extent of a two-dimensional surface or shape in the plane. Area can be understood as the amount of material with a given thickness that would be necessary to fashion a model of the shape, or the amount of paint necessary to cover the surface with a single coat...
. Thus for n = 2 the AM-GM inequality states that a square has the smallest perimeter among all rectangles with equal area.
The full inequality is an extension of this idea to n dimensions. Every vertex on an n-dimensional box is connected to n edges. If these edges are given lengths x1, x2, ..., xn, then x1 + x2 + … + xn is the total length of edges connected to the vertex. There are 2n vertices, so multiply this by 2n; however, since each edge meets two vertices, that counts every edge twice. Thus we divide by 2 and conclude that there are n2n−1 edges. There are equally many edges of each length and n lengths; hence there are 2n−1 edges of each length. The total edge-length is therefore 2n−1(x1 + x2 + … + xn). On the other hand,
is the total length of edges connected to a vertex on an n-dimensional cube of equal volume. Since the inequality says
we get
Thus the AM–GM inequality states that an n-cube
Hypercube
In geometry, a hypercube is an n-dimensional analogue of a square and a cube . It is a closed, compact, convex figure whose 1-skeleton consists of groups of opposite parallel line segments aligned in each of the space's dimensions, perpendicular to each other and of the same length.An...
has the smallest sum of lengths of edges connected to each vertex among all n-dimensional boxes with the same volume.
Generalizations
- There is a similar inequality for the weighted arithmetic mean and weighted geometric mean. Specifically, let the nonnegative numbers x1, x2, . . ., xn and the nonnegative weights α1, α2, . . ., αn be given. Set
. If α > 0, then the inequality
- holds with equality if and only if all the xk with αk > 0 are equal. Here the convention 00 = 1 is used.
- If all αk = 1, this reduces to the above AM–GM inequality.
Other generalizations of the inequality of arithmetic and geometric means include these:
- Muirhead's inequalityMuirhead's inequalityIn mathematics, Muirhead's inequality, named after Robert Franklin Muirhead, also known as the "bunching" method, generalizes the inequality of arithmetic and geometric means.-The "a-mean":For any real vectora=...
- Maclaurin's inequality
- generalized mean inequalityGeneralized meanIn mathematics, a generalized mean, also known as power mean or Hölder mean , is an abstraction of the Pythagorean means including arithmetic, geometric, and harmonic means.-Definition:...
Example application
Consider the following function:for x, y, and z all positive real numbers. Suppose we wish to find the minimum value of this function. Rewriting a bit, and applying the AM–GM inequality, we have:
 |
 |
 |
|
 |
|
 |
Further, we know that the two sides are equal exactly when all the terms of the mean are equal:
Proof by induction
There are several ways to prove the AM–GM inequality; for example, it can be inferred from Jensen's inequalityJensen's inequality
In mathematics, Jensen's inequality, named after the Danish mathematician Johan Jensen, relates the value of a convex function of an integral to the integral of the convex function. It was proved by Jensen in 1906. Given its generality, the inequality appears in many forms depending on the context,...
, using the concave function ln(x). It can also be proven using the rearrangement inequality. Considering length and required prerequisites, the proof by induction given below is probably the best recommendation for first reading.
With the arithmetic mean
of the non-negative real numbers x1,...,xn, the AM–GM statement is equivalent to
with equality if and only if μ = xi for all i = 1,...,n.
For the following proof we apply mathematical induction
Mathematical induction
Mathematical induction is a method of mathematical proof typically used to establish that a given statement is true of all natural numbers...
and only well-known rules of arithmetic.
Induction basis: For n = 1 the statement is true with equality.
Induction hypothesis: Suppose that the AM–GM statement holds for all choices of n non-negative real numbers.
Induction step: Consider n + 1 non-negative real numbers. Their arithmetic mean μ satisfies
If all numbers are equal to μ, then we have equality in the AM–GM statement and we are done. Otherwise we may find one number that is greater than μ and one that is smaller than μ, say xn > μ and xn+1 < μ. Then
Now consider the n numbers
with which are also non-negative. Since
μ is also the arithmetic mean of
and the induction hypothesis impliesDue to (*) we know that
hence
in particular μ > 0. Therefore, if at least one of the numbers x1,...,xn−1 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 get
which completes the proof.
Proof by Pólya
George PólyaGeorge Pólya
George Pólya was a Hungarian mathematician. He was a professor of mathematics from 1914 to 1940 at ETH Zürich and from 1940 to 1953 at Stanford University. He made fundamental contributions to combinatorics, number theory, numerical analysis and probability theory...
provided a proof similar to what follows. Let f(x) = ex−1 − x, with derivative f(x) = ex−1 − 1. Observe f(1) = 0 and hence that f has an absolute minimum of f(1) = 0. Now x ≤ ex−1 for all real x.
Consider a list of non-negative numbers
with arithmetic mean μ. By repeated application of the above inequality, we obtain the following:But the exponential argument can be simplified:
Returning to (1),
which produces the result:
Proof by Cauchy
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 CauchyAugustin Louis Cauchy
Baron Augustin-Louis Cauchy was a French mathematician who was an early pioneer of analysis. He started the project of formulating and proving the theorems of infinitesimal calculus in a rigorous manner, rejecting the heuristic principle of the generality of algebra exploited by earlier authors...
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 nx1, so their arithmetic mean is x1; and their product is x1n, so their geometric mean is x1; 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 n > 1.This case is significantly more complex, and we divide it into subcases.
The subcase where n = 2
If n = 2, then we have two terms, x1 and x2, and since (by our assumption) not all terms are equal, we have:
-
as desired.
The subcase where n
= 2k
Consider the case where n = 2k, where k is a positive integer. We proceed by mathematical induction.
In the base case, k = 1, so n = 2. We have already shown that the inequality holds where n = 2, so we are done.
Now, suppose that for a given k > 1, we have already shown that the inequality holds for n = 2k−1, and we wish to show that it holds for n = 2k. To do so, we proceed as follows:
-
where in the first inequality, the two sides are equal only if both of the following are true:
(in which case the first arithmetic mean and first geometric mean are both equal to x1, 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 2k 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 n is not a natural power of 2, then it is certainly less than some natural power of 2, since the sequence 2, 4, 8, . . ., 2k, . . . is unbounded above. Therefore, without loss of generality, let m be some natural power of 2 that is greater than n.
So, if we have n terms, then let us denote their arithmetic mean by α, and expand our list of terms thus:
We then have:
-
so
-
as desired.
Proof of the generalized AM–GM inequality using Jensen's inequality
Using the finite form of Jensen's inequality for the natural logarithm, we can prove the inequality between the weighted arithmetic mean and the weighted geometric mean stated above.
Since an xk with weight αk = 0 has no influence on the inequality, we may assume in the following that all weights are positive. If all xk are equal, then equality holds. Therefore, it remains to prove strict inequality if they are not all equal, which we will assume in the following, too. If at least one xk is zero (but not all), then the weighted geometric mean is zero, while the weighted arithmetic mean is positive, hence strict inequality holds. Therefore, we may assume also that all xk are positive.
Since the natural logarithm is strictly concaveConcave functionIn mathematics, a concave function is the negative of a convex function. A concave function is also synonymously called concave downwards, concave down, convex upwards, convex cap or upper convex.-Definition:...
, the finite form of Jensen's inequality and the functional equationFunctional equationIn mathematics, a functional equation is any equation that specifies a function in implicit form.Often, the equation relates the value of a function at some point with its values at other points. For instance, properties of functions can be determined by considering the types of functional...
s of the natural logarithm imply
Since the natural logarithm is strictly increasingMonotonic functionIn mathematics, a monotonic function is a function that preserves the given order. This concept first arose in calculus, and was later generalized to the more abstract setting of order theory....
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