AM–GM inequality explained

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 – i.e., with more than one variable – for two non-negative numbers and , is the statement that

(x + y)/(2) \ge \sqrt{xy} with equality if and only if .This case can be seen from the fact that the square of a real number is always non-negative (greater than or equal to zero) and from the elementary case of the binomial formula:\begin{align} 0 & \le (x - y) ^{2} \\ & = x ^{2-2xy+y} ^{2} \\ & = x ^{2+2xy+y} ^{2} - 4 xy \\ & = (x + y) ^{2} - 4 xy. \end{align} Hence, with equality precisely 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 , hence 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 are available to include weights or generalized means.

Background

The arithmetic mean, or less precisely the average, of a list of numbers is the sum of the numbers divided by :

(x _{1} + x _{2} + … + x _{n})/(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]{x _{1} ⋅ x _{2} … x _{n}.}

If, this is equal to the exponential of the arithmetic mean of the natural logarithms of the numbers:

\exp \left( (ln {x _{1})/(+) ln {x _{2}} + … + ln {x _{n}}{n}} \right).

Note: This does not apply exclusively to the exp function and natural logarithms. The base b of the exponentiation could be any positive real number except 1 if the logarithm is of base b.

The inequality

Restating the inequality using mathematical notation, we have that for any list of nonnegative real numbers,

(x _{1} + x _{2} + … + x _{n})/(n) \ge \sqrt[n]{x _{1} ⋅ x _{2} … x _{n},}

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,

\sqrt[n]{x _{1} x _{2} … x _{n}} 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.^[2]

Examples

Example 1

If

a,b,c> 0 , then the A.M.-G.M. tells us that( 1+ a)( 1+ b)( 1+ c)\ge 2 \sqrt{ 1 ⋅ {a}} ⋅ 2 \sqrt{ 1 ⋅ {b}} ⋅ 2 \sqrt{ 1 ⋅ {c}} = 8 \sqrt{abc}

Example 2

A simple upper bound for

n! can be found. AM-GM tells us1+2+...+ n \ge n\sqrt[n]{n!} (n(n +1 ))/(2) \ge n\sqrt[n]{n!}

and so

\left( (n +1)/(2) \right) ^{n} \ge n!

with equality at

n =1 .

Equivalently,

(n +1 ) ^{n} \ge 2 ^{nn!}

Example 3

Consider the function

f(x,y,z) = (x)/(y) + \sqrt{ (y)/(z) } + \sqrt[3]

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 ⋅ ((x)/(y) + (1)/(2) \sqrt{ (y)/(z))/(+) (1)/(\sqrt{ 2) (y)/(z) } + \frac \sqrt[3] + \frac \sqrt[3] + \frac \sqrt[3] }\\&=6\cdot\frac\endwithx _{1= (x)/(y)} , x _{2=x} \sqrt{ _{3= (1)/(2)} (y)/(z) },\qquad x_4=x_5=x_6=\frac \sqrt[3].

Applying the AM–GM inequality for, we get

\begin{align} f(x,y,z) &\ge 6 ⋅ \sqrt[ 6 ]{ (x)/(y) ⋅ (1)/(\sqrt{ 2) (y)/(z) } \cdot \frac \sqrt \cdot \frac \sqrt[3] \cdot \frac \sqrt[3] \cdot \frac \sqrt[3] }\\&= 6 \cdot \sqrt[6]\\&= 2^ \cdot 3^.\end

Further, we know that the two sides are equal exactly when all the terms of the mean are equal:

f(x,y,z) = 2 ^{2/3} ⋅ 3 ^{1/2} when (x)/(y) = (1)/(\sqrt{ 2) (y)/(z) } = \frac \sqrt[3].

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)/() }\,t\biggr)\quad\mbox\quad t>0.

Applications

An important practical application in financial mathematics is to computing the rate of return: the annualized return, computed via the geometric mean, is less than the average annual return, computed by the arithmetic mean (or equal if all returns are equal). This is important in analyzing investments, as the average return overstates the cumulative effect. It can also be used to prove the Cauchy–Schwarz inequality.

Proofs of the AM–GM inequality

The AM–GM inequality is also known for the variety of methods that can be used to prove it.

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

log \left( (\sum _{i} x _{i})/(n) \right) \geq \sum (1)/(n) log x _{i} = \sum \left( log x ^{1 /n} _{i} \right) = log \left( \prod x ^{1 /n} _{i} \right).

Taking antilogs 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

\alpha = (x _{1+x} _{2+ … +x} _{n})/(n) \ge \sqrt[n]{x _{1x} _{2} … x _{n}=\beta}

with equality only when all numbers are equal.

If not all numbers are equal, then there exist

x _{i,x} _{j} such that x _{i<\alpha<x} _{j} . Replacing by \alpha and by (x _{i+x} _{j-\alpha)} will leave the arithmetic mean of the numbers unchanged, but will increase the geometric mean because\alpha(x _{j+x} _{i-\alpha)-x} _{ix} _{j=(\alpha-x} _{i)(x} _{j-\alpha)>0}

If the numbers are still not equal, we continue replacing numbers as above. After at most

(n -1 ) such replacement steps all the numbers will have been replaced with \alpha while the geometric mean strictly increases at each step. After the last step, the geometric mean will be \sqrt[n]{\alpha\alpha … \alpha} = \alpha , 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

x _{i,x} _{j} such that 0 <x _{i<\beta<x} _{j} . Replacing x _{i} by \beta and x _{j} by (x _{ix} _{j})/(\beta) leaves the geometric mean unchanged but strictly decreases the arithmetic mean sincex _{i} + x _{j} - \beta - (x _{i} x _{j})/(\beta) = ((\beta - x _{i)(x} _{j} - \beta))/(\beta) > 0 . 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

\alpha ^{n\ge} x _{1} x _{2} … x _{n} with equality if and only if for all

Notes and References

1. Web site: Euclid's Elements, Book V, Proposition 25 .
2. Book: Steele , J. Michael . The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities. Cambridge University Press. MAA Problem Books Series. 2004. 978-0-521-54677-5. 54079548.