AM–GM inequality

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.

The simplest non-trivial case — i.e., with more than one variable — for two non-negative numbers $x$ and $y$ , is the statement that

{\frac {x+y}{2}}\geq {\sqrt {xy}}

with equality if and only if $x = y$ . 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 $(a \pm b) 2 = a 2 \pm 2 ab + b 2$ of the binomial formula:

{\begin{aligned}0&\leq (x-y)^{2}\\&=x^{2}-2xy+y^{2}\\&=x^{2}+2xy+y^{2}-4xy\\&=(x+y)^{2}-4xy.\end{aligned}}

Hence $(x + y) 2 \geq 4 xy$ , with equality precisely when $(x - y) 2 = 0$ , i.e. $x = y$ . The AM–GM inequality then follows from taking the positive square root of both sides.

For a geometrical interpretation, consider a rectangle with sides of length $x$ and $y$ , hence it has perimeter $2 x + 2 y$ and area $xy$ . Similarly, a square with all sides of length $\sqrt xy$ has the perimeter $4 \sqrt xy$ and the same area as the rectangle. The simplest non-trivial case of the AM–GM inequality implies for the perimeters that $2 x + 2 y \geq 4 \sqrt xy$ and that only the square has the smallest perimeter amongst all rectangles of equal area.

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 $n$ numbers $x 1, x 2, . . . , x n$ is the sum of the numbers divided by $n$ :

{\frac {x_{1}+x_{2}+\cdots +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}\cdot x_{2}\cdots x_{n}}}.

If $x 1, x 2, . . . , x n > 0$ , this is equal to the exponential of the arithmetic mean of the natural logarithms of the numbers:

\exp \left({\frac {\ln {x_{1}}+\ln {x_{2}}+\cdots +\ln {x_{n}}}{n}}\right).

The inequality

Restating the inequality using mathematical notation, we have that for any list of $n$ nonnegative real numbers $x 1, x 2, . . . , x n$ ,

{\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}\geq {\sqrt[{n}]{x_{1}\cdot x_{2}\cdots x_{n}}}\,,

and that equality holds if and only if $x 1 = x 2 = \cdot \cdot \cdot = x n$ .

Geometric interpretation

In two dimensions, $2 x 1 + 2 x 2$ is the perimeter of a rectangle with sides of length $x 1$ and $x 2$ . Similarly, $4 \sqrt x 1 x 2$ is the perimeter of a square with the same area, $x 1 x 2$ , as that rectangle. Thus for $n = 2$ 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 $n$ dimensions. Every vertex of an $n$ -dimensional box is connected to $n$ edges. If these edges' lengths are $x 1, x 2, . . . , x n$ , then $x 1 + x 2 + \cdot \cdot \cdot + x n$ is the total length of edges incident to the vertex. There are $2 n$ vertices, so we multiply this by $2 n$ ; since each edge, however, meets two vertices, every edge is counted twice. Therefore, we divide by $2$ and conclude that there are $2 n -1 n$ edges. There are equally many edges of each length and $n$ lengths; hence there are $2 n -1$ edges of each length and the total of all edge lengths is $2 n -1 (x 1 + x 2 + \cdot \cdot \cdot + x n)$ . On the other hand,

2^{n-1}(x_{1}+\ldots +x_{n})=2^{n-1}n{\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}

is the total length of edges connected to a vertex on an $n$ -dimensional cube of equal volume, since in this case $x 1 =...= x n$ . Since the inequality says

{x_{1}+x_{2}+\cdots +x_{n} \over n}\geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}},

it can be restated by multiplying through by $n 2 n -1$ to obtain

2^{n-1}(x_{1}+x_{2}+\cdots +x_{n})\geq 2^{n-1}n{\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}

with equality if and only if $x 1 = x 2 = \cdot \cdot \cdot = x n$ .

Thus the AM–GM inequality states that only the $n$ -cube has the smallest sum of lengths of edges connected to each vertex amongst all $n$ -dimensional boxes with the same volume.^[2]

Example application

Consider the function

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

for all positive real numbers $x$ , $y$ and $z$ . Suppose we wish to find the minimal value of this function. First we rewrite it a bit:

{\begin{aligned}f(x,y,z)&=6\cdot {\frac {{\frac {x}{y}}+{\frac {1}{2}}{\sqrt {\frac {y}{z}}}+{\frac {1}{2}}{\sqrt {\frac {y}{z}}}+{\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}+{\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}+{\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}}{6}}\\&=6\cdot {\frac {x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}}{6}}\end{aligned}}

with

x_{1}={\frac {x}{y}},\qquad x_{2}=x_{3}={\frac {1}{2}}{\sqrt {\frac {y}{z}}},\qquad x_{4}=x_{5}=x_{6}={\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}.

Applying the AM–GM inequality for $n = 6$ , we get

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

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}\cdot 3^{1/2}\quad {\mbox{when}}\quad {\frac {x}{y}}={\frac {1}{2}}{\sqrt {\frac {y}{z}}}={\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}.

All the points $(x, y, z)$ satisfying these conditions lie on a half-line starting at the origin and are given by

(x,y,z)={\biggr (}t,{\sqrt[{3}]{2}}{\sqrt {3}}\,t,{\frac {3{\sqrt {3}}}{2}}\,t{\biggr )}\quad {\mbox{with}}\quad t>0.

Practical 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.

Proofs of the AM–GM inequality

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

Taking antilogs of the far left and far right sides, we have the AM–GM inequality.

Proofs by induction

We have to show that

{\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}\geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}

with equality only when all numbers are equal. If $x i \neq x j$ , then replacing both $x i$ and $x j$ by $(x i + x j)/2$ will leave the arithmetic mean on the left-hand side unchanged, but will increase the geometric mean on the right-hand side because

{\Bigl (}{\frac {x_{i}+x_{j}}{2}}{\Bigr )}^{2}-x_{i}x_{j}={\Bigl (}{\frac {x_{i}-x_{j}}{2}}{\Bigr )}^{2}>0.

Thus the right-hand side will be largest when all $x i$ s are equal to the arithmetic mean

\alpha ={\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}},

thus as this is then the largest value of right-hand side of the expression, we have

{\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}=\alpha ={\sqrt[{n}]{\alpha \alpha \cdots \alpha }}\geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}.

This is a valid proof for the case $n = 2$ , but the procedure of taking iteratively pairwise averages may fail to produce $n$ equal numbers in the case $n \geq 3$ . An example of this case is $x 1 = x 2 \neq x 3$ : Averaging two different numbers produces two equal numbers, but the third one is still different. Therefore, we never actually get an inequality involving the geometric mean of three equal numbers.

Hence, an additional trick or a modified argument is necessary to turn the above idea into a valid proof for the case $n \geq 3$ .

Proof by induction #1

Of the non-negative real numbers $x 1, . . . , x n$ , the AM–GM statement is equivalent to

\alpha ^{n}\geq x_{1}x_{2}\cdots x_{n}

with equality if and only if $α = x i$ for all $i \in {1, . . . , n}$ .

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 $x 1, . . . , x n +1$ , . Their arithmetic mean $α$ satisfies

(n+1)\alpha =\ x_{1}+\cdots +x_{n}+x_{n+1}.

If all the $x i$ are equal to $α$ , then we have equality in the AM–GM statement and we are done. In the case where some are not equal to $α$ , there must exist one number that is greater than the arithmetic mean $α$ , and one that is smaller than $α$ . Without loss of generality, we can reorder our $x i$ in order to place these two particular elements at the end: $x n > α$ and $x n +1 < α$ . Then

x_{n}-\alpha >0\qquad \alpha -x_{n+1}>0

\implies (x_{n}-\alpha )(\alpha -x_{n+1})>0\,.\qquad (*)

Now define $y$ with

y:=x_{n}+x_{n+1}-\alpha \geq x_{n}-\alpha >0\,,

and consider the $n$ numbers $x 1, . . . , x n -1, y$ which are all non-negative. Since

(n+1)\alpha =x_{1}+\cdots +x_{n-1}+x_{n}+x_{n+1}

n\alpha =x_{1}+\cdots +x_{n-1}+\underbrace {x_{n}+x_{n+1}-\alpha } _{=\,y},

Thus, $α$ is also the arithmetic mean of $n$ numbers $x 1, . . . , x n -1, y$ and the induction hypothesis implies

\alpha ^{n+1}=\alpha ^{n}\cdot \alpha \geq x_{1}x_{2}\cdots x_{n-1}y\cdot \alpha .\qquad (**)

Due to (*) we know that

(\underbrace {x_{n}+x_{n+1}-\alpha } _{=\,y})\alpha -x_{n}x_{n+1}=(x_{n}-\alpha )(\alpha -x_{n+1})>0,

hence

y\alpha >x_{n}x_{n+1}\,,\qquad ({*}{*}{*})

in particular $α > 0$ . Therefore, if at least one of the numbers $x 1, . . . , x n -1$ 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 can substitute (***) into (**) to get

\alpha ^{n+1}>x_{1}x_{2}\cdots x_{n-1}x_{n}x_{n+1}\,,

which completes the proof.

Proof by induction #2

First of all we shall prove that for real numbers $x 1 < 1$ and $x 2 > 1$ there follows

x_{1}+x_{2}>x_{1}x_{2}+1.

Indeed, multiplying both sides of the inequality $x 2 > 1$ by $1 - x 1$ , gives

x_{2}-x_{1}x_{2}>1-x_{1},

whence the required inequality is obtained immediately.

Now, we are going to prove that for positive real numbers $x 1, . . . , x n$ satisfying $x 1 . . . x n = 1$ , there holds

x_{1}+\cdots +x_{n}\geq n.

The equality holds only if $x 1 = ... = x n = 1$ .

Induction basis: For $n = 2$ the statement is true because of the above property.

Induction hypothesis: Suppose that the statement is true for all natural numbers up to $n - 1$ .

Induction step: Consider natural number $n$ , i.e. for positive real numbers $x 1, . . . , x n$ , there holds $x 1 . . . x n = 1$ . There exists at least one $x k < 1$ , so there must be at least one $x j > 1$ . Without loss of generality, we let $k = n - 1$ and $j = n$ .

Further, the equality $x 1 . . . x n = 1$ we shall write in the form of $(x 1 . . . x n -2) (x n -1 x n) = 1$ . Then, the induction hypothesis implies

(x_{1}+\cdots +x_{n-2})+(x_{n-1}x_{n})>n-1.

However, taking into account the induction basis, we have

{\begin{aligned}x_{1}+\cdots +x_{n-2}+x_{n-1}+x_{n}&=(x_{1}+\cdots +x_{n-2})+(x_{n-1}+x_{n})\\&>(x_{1}+\cdots +x_{n-2})+x_{n-1}x_{n}+1\\&>n,\end{aligned}}

which completes the proof.

For positive real numbers $a 1, . . . , a n$ , let's denote

x_{1}={\frac {a_{1}}{\sqrt[{n}]{a_{1}\cdots a_{n}}}},...,x_{n}={\frac {a_{n}}{\sqrt[{n}]{a_{1}\cdots a_{n}}}}.

The numbers $x 1, . . . , x n$ satisfy the condition $x 1 . . . x n = 1$ . So we have

{\frac {a_{1}}{\sqrt[{n}]{a_{1}\cdots a_{n}}}}+\cdots +{\frac {a_{n}}{\sqrt[{n}]{a_{1}\cdots a_{n}}}}\geq n,

whence we obtain

{\frac {a_{1}+\cdots +a_{n}}{n}}\geq {\sqrt[{n}]{a_{1}\cdots a_{n}}},

with the equality holding only for $a 1 = ... = a n = 1$ .

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.^[3]

The case where all the terms are equal

If all the terms are equal:

x_{1}=x_{2}=\cdots =x_{n},

then their sum is $nx 1$ , so their arithmetic mean is $x 1$ ; and their product is $x 1 n$ , so their geometric mean is $x 1$ ; 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, $x 1$ and $x 2$ , and since (by our assumption) not all terms are equal, we have:

{\begin{aligned}{\Bigl (}{\frac {x_{1}+x_{2}}{2}}{\Bigr )}^{2}-x_{1}x_{2}&={\frac {1}{4}}(x_{1}^{2}+2x_{1}x_{2}+x_{2}^{2})-x_{1}x_{2}\\&={\frac {1}{4}}(x_{1}^{2}-2x_{1}x_{2}+x_{2}^{2})\\&={\Bigl (}{\frac {x_{1}-x_{2}}{2}}{\Bigr )}^{2}>0,\end{aligned}}

hence

{\frac {x_{1}+x_{2}}{2}}>{\sqrt {x_{1}x_{2}}}

as desired.

The subcase where n = 2^k

Consider the case where $n = 2 k$ , 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 when $n = 2$ , so we are done.

Now, suppose that for a given $k > 1$ , we have already shown that the inequality holds for $n = 2 k -1$ , and we wish to show that it holds for $n = 2 k$ . To do so, we apply the inequality twice for $2 k -1$ numbers and once for $2$ numbers to obtain:

{\begin{aligned}{\frac {x_{1}+x_{2}+\cdots +x_{2^{k}}}{2^{k}}}&{}={\frac {{\frac {x_{1}+x_{2}+\cdots +x_{2^{k-1}}}{2^{k-1}}}+{\frac {x_{2^{k-1}+1}+x_{2^{k-1}+2}+\cdots +x_{2^{k}}}{2^{k-1}}}}{2}}\\[7pt]&\geq {\frac {{\sqrt[{2^{k-1}}]{x_{1}x_{2}\cdots x_{2^{k-1}}}}+{\sqrt[{2^{k-1}}]{x_{2^{k-1}+1}x_{2^{k-1}+2}\cdots x_{2^{k}}}}}{2}}\\[7pt]&\geq {\sqrt {{\sqrt[{2^{k-1}}]{x_{1}x_{2}\cdots x_{2^{k-1}}}}{\sqrt[{2^{k-1}}]{x_{2^{k-1}+1}x_{2^{k-1}+2}\cdots x_{2^{k}}}}}}\\[7pt]&={\sqrt[{2^{k}}]{x_{1}x_{2}\cdots x_{2^{k}}}}\end{aligned}}

where in the first inequality, the two sides are equal only if

x_{1}=x_{2}=\cdots =x_{2^{k-1}}

and

x_{2^{k-1}+1}=x_{2^{k-1}+2}=\cdots =x_{2^{k}}

(in which case the first arithmetic mean and first geometric mean are both equal to $x 1$ , 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 $2 k$ numbers are equal, it is not possible for both inequalities to be equalities, so we know that:

{\frac {x_{1}+x_{2}+\cdots +x_{2^{k}}}{2^{k}}}>{\sqrt[{2^{k}}]{x_{1}x_{2}\cdots x_{2^{k}}}}

as desired.

The subcase where n < 2^k

If $n$ is not a natural power of $2$ , then it is certainly less than some natural power of 2, since the sequence $2, 4, 8, . . . , 2 k, . . .$ 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:

x_{n+1}=x_{n+2}=\cdots =x_{m}=\alpha .

We then have:

{\begin{aligned}\alpha &={\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}\\[6pt]&={\frac {{\frac {m}{n}}\left(x_{1}+x_{2}+\cdots +x_{n}\right)}{m}}\\[6pt]&={\frac {x_{1}+x_{2}+\cdots +x_{n}+{\frac {m-n}{n}}\left(x_{1}+x_{2}+\cdots +x_{n}\right)}{m}}\\[6pt]&={\frac {x_{1}+x_{2}+\cdots +x_{n}+\left(m-n\right)\alpha }{m}}\\[6pt]&={\frac {x_{1}+x_{2}+\cdots +x_{n}+x_{n+1}+\cdots +x_{m}}{m}}\\[6pt]&>{\sqrt[{m}]{x_{1}x_{2}\cdots x_{n}x_{n+1}\cdots x_{m}}}\\[6pt]&={\sqrt[{m}]{x_{1}x_{2}\cdots x_{n}\alpha ^{m-n}}}\,,\end{aligned}}

so

\alpha ^{m}>x_{1}x_{2}\cdots x_{n}\alpha ^{m-n}

and

\alpha >{\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}

as desired.

Proof by induction using basic calculus

The following proof uses mathematical induction and some basic differential calculus.

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: In order to prove the statement for $n + 1$ non-negative real numbers $x 1, . . . , x n, x n +1$ , we need to prove that

{\frac {x_{1}+\cdots +x_{n}+x_{n+1}}{n+1}}-({x_{1}\cdots x_{n}x_{n+1}})^{\frac {1}{n+1}}\geq 0

with equality only if all the $n + 1$ 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 $n + 1$ numbers are positive.

We consider the last number $x n +1$ as a variable and define the function

f(t)={\frac {x_{1}+\cdots +x_{n}+t}{n+1}}-({x_{1}\cdots x_{n}t})^{\frac {1}{n+1}},\qquad t>0.

Proving the induction step is equivalent to showing that $f (t) \geq 0$ for all $t > 0$ , with $f (t) = 0$ only if $x 1, . . . , x n$ and $t$ are all equal. This can be done by analyzing the critical points of $f$ using some basic calculus.

The first derivative of $f$ is given by

f'(t)={\frac {1}{n+1}}-{\frac {1}{n+1}}({x_{1}\cdots x_{n}})^{\frac {1}{n+1}}t^{-{\frac {n}{n+1}}},\qquad t>0.

A critical point $t 0$ has to satisfy $f' (t 0) = 0$ , which means

({x_{1}\cdots x_{n}})^{\frac {1}{n+1}}t_{0}^{-{\frac {n}{n+1}}}=1.

After a small rearrangement we get

t_{0}^{\frac {n}{n+1}}=({x_{1}\cdots x_{n}})^{\frac {1}{n+1}},

and finally

t_{0}=({x_{1}\cdots x_{n}})^{\frac {1}{n}},

which is the geometric mean of $x 1, . . . , x n$ . This is the only critical point of $f$ . Since $f'' (t) > 0$ for all $t > 0$ , the function $f$ is strictly convex and has a strict global minimum at $t 0$ . Next we compute the value of the function at this global minimum:

{\begin{aligned}f(t_{0})&={\frac {x_{1}+\cdots +x_{n}+({x_{1}\cdots x_{n}})^{1/n}}{n+1}}-({x_{1}\cdots x_{n}})^{\frac {1}{n+1}}({x_{1}\cdots x_{n}})^{\frac {1}{n(n+1)}}\\&={\frac {x_{1}+\cdots +x_{n}}{n+1}}+{\frac {1}{n+1}}({x_{1}\cdots x_{n}})^{\frac {1}{n}}-({x_{1}\cdots x_{n}})^{\frac {1}{n}}\\&={\frac {x_{1}+\cdots +x_{n}}{n+1}}-{\frac {n}{n+1}}({x_{1}\cdots x_{n}})^{\frac {1}{n}}\\&={\frac {n}{n+1}}{\Bigl (}{\frac {x_{1}+\cdots +x_{n}}{n}}-({x_{1}\cdots x_{n}})^{\frac {1}{n}}{\Bigr )}\\&\geq 0,\end{aligned}}

where the final inequality holds due to the induction hypothesis. The hypothesis also says that we can have equality only when $x 1, . . . , x n$ are all equal. In this case, their geometric mean $t 0$ has the same value, Hence, unless $x 1, . . . , x n, x n +1$ are all equal, we have $f (x n +1) > 0$ . This completes the proof.

This technique can be used in the same manner to prove the generalized AM–GM inequality and Cauchy–Schwarz inequality in Euclidean space $R n$ .

Proof by Pólya using the exponential function

George Pólya provided a proof similar to what follows. Let $f (x) = e x -1 - x$ for all real $x$ , with first derivative $f' (x) = e x -1 - 1$ and second derivative $f'' (x) = e x -1$ . Observe that $f (1) = 0$ , $f' (1) = 0$ and $f'' (x) > 0$ for all real $x$ , hence $f$ is strictly convex with the absolute minimum at $x = 1$ . Hence $x \leq e x -1$ for all real $x$ with equality only for $x = 1$ .

Consider a list of non-negative real numbers $x 1, x 2, . . . , x n$ . If they are all zero, then the AM–GM inequality holds with equality. Hence we may assume in the following for their arithmetic mean $α > 0$ . By $n$ -fold application of the above inequality, we obtain that

{\begin{aligned}{{\frac {x_{1}}{\alpha }}{\frac {x_{2}}{\alpha }}\cdots {\frac {x_{n}}{\alpha }}}&\leq {e^{{\frac {x_{1}}{\alpha }}-1}e^{{\frac {x_{2}}{\alpha }}-1}\cdots e^{{\frac {x_{n}}{\alpha }}-1}}\\&=\exp {\Bigl (}{\frac {x_{1}}{\alpha }}-1+{\frac {x_{2}}{\alpha }}-1+\cdots +{\frac {x_{n}}{\alpha }}-1{\Bigr )},\qquad (*)\end{aligned}}

with equality if and only if $x i = α$ for every $i \in {1, . . . , n}$ . The argument of the exponential function can be simplified:

{\begin{aligned}{\frac {x_{1}}{\alpha }}-1+{\frac {x_{2}}{\alpha }}-1+\cdots +{\frac {x_{n}}{\alpha }}-1&={\frac {x_{1}+x_{2}+\cdots +x_{n}}{\alpha }}-n\\&={\frac {n\alpha }{\alpha }}-n\\&=0.\end{aligned}}

Returning to $(*)$ ,

{\frac {x_{1}x_{2}\cdots x_{n}}{\alpha ^{n}}}\leq e^{0}=1,

which produces $x 1 x 2 \cdot \cdot \cdot x n \leq α n$ , hence the result^[4]

{\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}\leq \alpha .

Proof by Lagrangian Multipliers

If any of the $x_{i}$ are $0$ , then there is nothing to prove. So we may assume all the $x_{i}$ are strictly positive.

Because the arithmetic and geometric means are homogeneous of degree 1, without loss of generality assume that $\prod _{i=1}^{n}x_{i}=1$ . Set $G(x_{1},x_{2},\ldots ,x_{n})=\prod _{i=1}^{n}x_{i}$ , and $F(x_{1},x_{2},\ldots ,x_{n})={\frac {1}{n}}\sum _{i=1}^{n}x_{i}$ . The inequality will be proved (together with the equality case) if we can show that the minimum of $F(x_{1},x_{2},...,x_{n}),$ subject to the constraint $G(x_{1},x_{2},\ldots ,x_{n})=1,$ is equal to $1$ , and the minimum is only achieved when $x_{1}=x_{2}=\cdots =x_{n}=1$ . Let us first show that the constraint problem has a global minimum.

Set $K=\{(x_{1},x_{2},\ldots ,x_{n})\colon x_{1},x_{2},\ldots ,x_{n}\leq n\}$ . Since the intersection $K\cap \{G=1\}$ is compact, the extreme value theorem guarantees that the minimum of $F(x_{1},x_{2},...,x_{n})$ subject to the constraints $G(x_{1},x_{2},\ldots ,x_{n})=1$ and $(x_{1},x_{2},\ldots ,x_{n})\in K$ is attained at some point inside $K$ . On the other hand, observe that if any of the $x_{i}>n$ , then $F(x_{1},x_{2},\ldots ,x_{n})>1$ , while $F(1,1,\ldots ,1)=1$ , and $(1,1,\ldots ,1)\in K\cap \{G=1\}$ . This means that the minimum inside $K\cap \{G=1\}$ is in fact a global minimum, since the value of $F$ at any point inside $K\cap \{G=1\}$ is certainly no smaller than the minimum, and the value of $F$ at any point $(y_{1},y_{2},\ldots ,y_{n})$ not inside $K$ is strictly bigger than the value at $(1,1,\ldots ,1)$ , which is no smaller than the minimum.

The method of Lagrange multipliers says that the global minimum is attained at a point $(x_{1},x_{2},\ldots ,x_{n})$ where the gradient of $F(x_{1},x_{2},\ldots ,x_{n})$ is $\lambda$ times the gradient of $G(x_{1},x_{2},\ldots ,x_{n})$ , for some $\lambda$ . We will show that the only point at which this happens is when $x_{1}=x_{2}=\cdots =x_{n}=1$ and $F(x_{1},x_{2},...,x_{n})=1.$

Compute ${\frac {\partial F}{\partial x_{i}}}={\frac {1}{n}}$ and

${\frac {\partial G}{\partial x_{i}}}=\prod _{j\neq i}x_{j}={\frac {G(x_{1},x_{2},\ldots ,x_{n})}{x_{i}}}={\frac {1}{x_{i}}}$

along the constraint. Setting the gradients proportional to one another therefore gives for each $i$ that ${\frac {1}{n}}={\frac {\lambda }{x_{i}}},$ and so $n\lambda =x_{i}.$ Since the left-hand side does not depend on $i$ , it follows that $x_{1}=x_{2}=\cdots =x_{n}$ , and since $G(x_{1},x_{2},\ldots ,x_{n})=1$ , it follows that $x_{1}=x_{2}=\cdots =x_{n}=1$ and $F(x_{1},x_{2},\ldots ,x_{n})=1$ , as desired.

Generalizations

Weighted AM–GM inequality

There is a similar inequality for the weighted arithmetic mean and weighted geometric mean. Specifically, let the nonnegative numbers $x 1, x 2, . . . , x n$ and the nonnegative weights $w 1, w 2, . . . , w n$ be given. Set $w = w 1 + w 2 + \cdot \cdot \cdot + w n$ . If $w > 0$ , then the inequality

{\frac {w_{1}x_{1}+w_{2}x_{2}+\cdots +w_{n}x_{n}}{w}}\geq {\sqrt[{w}]{x_{1}^{w_{1}}x_{2}^{w_{2}}\cdots x_{n}^{w_{n}}}}

holds with equality if and only if all the $x k$ with $w k > 0$ are equal. Here the convention $00 = 1$ is used.

If all $w k = 1$ , this reduces to the above inequality of arithmetic and geometric means.

Proof 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 $x k$ with weight $w k = 0$ has no influence on the inequality, we may assume in the following that all weights are positive. If all $x k$ 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 $x k$ 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 $x k$ are positive.

Since the natural logarithm is strictly concave, the finite form of Jensen's inequality and the functional equations of the natural logarithm imply

{\begin{aligned}\ln {\Bigl (}{\frac {w_{1}x_{1}+\cdots +w_{n}x_{n}}{w}}{\Bigr )}&>{\frac {w_{1}}{w}}\ln x_{1}+\cdots +{\frac {w_{n}}{w}}\ln x_{n}\\&=\ln {\sqrt[{w}]{x_{1}^{w_{1}}x_{2}^{w_{2}}\cdots x_{n}^{w_{n}}}}.\end{aligned}}

Since the natural logarithm is strictly increasing,

{\frac {w_{1}x_{1}+\cdots +w_{n}x_{n}}{w}}>{\sqrt[{w}]{x_{1}^{w_{1}}x_{2}^{w_{2}}\cdots x_{n}^{w_{n}}}}.

Other generalizations

Other generalizations of the inequality of arithmetic and geometric means include:

Notes

References

^ Hoffman, D. G. (1981), "Packing problems and inequalities", in Klarner, David A. (ed.), The Mathematical Gardner, Springer, pp. 212–225, doi:10.1007/978-1-4684-6686-7_19
^ Steele, J. Michael (2004). The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities. MAA Problem Books Series. Cambridge University Press. ISBN 978-0-521-54677-5. OCLC 54079548.
^ Cauchy, Augustin-Louis (1821). Cours d'analyse de l'École Royale Polytechnique, première partie, Analyse algébrique, Paris. The proof of the inequality of arithmetic and geometric means can be found on pages 457ff.
^ Arnold, Denise; Arnold, Graham (1993). Four unit mathematics. Hodder Arnold H&S. p. 242. ISBN 978-0-340-54335-1. OCLC 38328013.

External links

Arthur Lohwater (1982). "Introduction to Inequalities". Online e-book in PDF format.

[1] Hoffman, D. G. (1981), "Packing problems and inequalities", in Klarner, David A. (ed.), The Mathematical Gardner, Springer, pp. 212–225, doi:10.1007/978-1-4684-6686-7_19

[2] Steele, J. Michael (2004). The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities. MAA Problem Books Series. Cambridge University Press. ISBN 978-0-521-54677-5. OCLC 54079548.

[3] Cauchy, Augustin-Louis (1821). Cours d'analyse de l'École Royale Polytechnique, première partie, Analyse algébrique, Paris. The proof of the inequality of arithmetic and geometric means can be found on pages 457ff.

[4] Arnold, Denise; Arnold, Graham (1993). Four unit mathematics. Hodder Arnold H&S. p. 242. ISBN 978-0-340-54335-1. OCLC 38328013.

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Background

The inequality

Geometric interpretation

Example application

Practical applications

Proofs of the AM–GM inequality

Proof using Jensen's inequality

Proofs by induction

Proof by induction #1

Proof by induction #2

Proof by Cauchy using forward–backward induction

The case where all the terms are equal

The case where not all the terms are equal

The subcase where n = 2

The subcase where n = 2k

The subcase where n < 2k

Proof by induction using basic calculus

Proof by Pólya using the exponential function

Proof by Lagrangian Multipliers

Generalizations

Weighted AM–GM inequality

Proof using Jensen's inequality

Other generalizations

See also

Notes

References

External links

The subcase where n = 2^k

The subcase where n < 2^k