AM–GM inequality

The inequality

Restating the inequality using mathematical notation, we have that for any list of n nonnegative real numbers x₁, x₂, . . . , x_n,

${\displaystyle \frac{x_1+x_2+\cdots +x_n}{n}\ge \sqrt[n]{x_1\cdot x_2\cdots x_n},}$

and that equality holds if and only if x₁ = x₂ = · · · = x_n.

Geometric interpretation

In two dimensions, 2x₁ + 2x₂ is the perimeter of a rectangle with sides of length x₁ and x₂. Similarly, 4√x₁x₂ is the perimeter of a square with the same area, x₁x₂, 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. Consider an n-dimensional box with edge lengths x₁, x₂, . . . , x_n. Every vertex of the box is connected to n edges of different directions, so the average length of edges incident to the vertex is (x₁ + x₂ + · · · + x_n)/n. On the other hand, ${\displaystyle \sqrt[n]{x_1{x}_2\cdots x_n}}$ is the edge length of an n-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 n-cube has the smallest average length of edges connected to each vertex amongst all n-dimensional boxes with the same volume.

Examples

Example 1

If ${\displaystyle a,b,c>0}$, then the AM-GM inequality tells us that

${\displaystyle (1+a)(1+b)(1+c)\ge 2\sqrt{1\cdot a}\cdot 2\sqrt{1\cdot b}\cdot 2\sqrt{1\cdot c}=8\sqrt{abc}}$

Example 2

A simple upper bound for ${\displaystyle n!}$ can be found. AM-GM tells us

${\displaystyle 1+2+\cdots +n\ge n\sqrt[n]{n!}}$

${\displaystyle \frac{n(n+1)}{2}\ge n\sqrt[n]{n!}}$

and so

${\displaystyle {\left(\frac{n+1}{2}\right)}^n\ge n!}$

with equality at ${\displaystyle n=1}$.

Equivalently,

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

Example 3

Consider the function

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

for all positive real numbers x, y and z. Suppose we wish to find the minimal value of this function. It can be rewritten as:

${\displaystyle \begin{array}{cc}f(x,y,z)& =6\cdot \frac{\frac{x}{y}+\frac{1}{2}\sqrt{\frac{y}{z}}+\frac{1}{2}\sqrt{\frac{y}{z}}+\frac{1}{3}\sqrt[3]{\frac{z}{x}}+\frac{1}{3}\sqrt[3]{\frac{z}{x}}+\frac{1}{3}\sqrt[3]{\frac{z}{x}}}{6}\\ & =6\cdot \frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}\end{array}}$

with

${\displaystyle x_1=\frac{x}{y},x_2=x_3=\frac{1}{2}\sqrt{\frac{y}{z}},x_4=x_5=x_6=\frac{1}{3}\sqrt[3]{\frac{z}{x}}.}$

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

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

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

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

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

${\displaystyle (x,y,z)=(t,\sqrt[3]{2}\sqrt{3}t,\frac{3\sqrt{3}}{2}t)\text{with}t>0.}$

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 ${\displaystyle n}$ vertices is bounded from above by ${\displaystyle \frac{n^2}{4}}$. The vertices of a bipartite graph with ${\displaystyle n}$ vertices may be divided into two classes ${\displaystyle A}$ and ${\displaystyle B}$. Since we can only draw edges between vertices in one class to another, the maximum number of edges is given by ${\displaystyle |A||B|}$. By the AM-GM inequality, we can bound ${\displaystyle |A||B|}$ from above like so:

${\displaystyle |A||B|\le (\frac{|A|+|B|}{2}{)}^2\le \frac{n^2}{4}}$

Nonnegative polynomials

The Motzkin polynomial $${\displaystyle x^4{y}^2+x^2{y}^4-3x^2{y}^2+1}$$ is a nonnegative polynomial which is not a sum of square polynomials. It can be proven nonnegative using the AM-GM inequality with ${\displaystyle x_1=x^4{y}^2}$, ${\displaystyle x_2=x^2{y}^4}$, and ${\displaystyle x_3=1}$, that is, $${\displaystyle \sqrt[3]{(x^4{y}^2)\cdot (x^2{y}^4)\cdot (1)}\le \frac{(x^4{y}^2)+(x^2{y}^4)+(1)}{3}.}$$ Simplifying and multiplying both sides by 3 gives $${\displaystyle 3x^2{y}^2\le x^4{y}^2+x^2{y}^4+1,}$$ so $${\displaystyle 0\le x^4{y}^2+x^2{y}^4-3x^2{y}^2+1.}$$

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

${\displaystyle \log \left(\frac{\sum x_i}{n}\right)\ge \frac{1}{n}\sum \log x_i=\frac{1}{n}\log \left(\prod x_i\right)=\log \left({\left(\prod x_i\right)}^{1/n}\right).}$

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

${\displaystyle \alpha =\frac{x_1+x_2+\cdots +x_n}{n}\ge \sqrt[n]{x_1{x}_2\cdots x_n}=\beta }$

with equality only when all numbers are equal.

If not all numbers are equal, then there exist ${\displaystyle x_i,x_j}$ such that ${\displaystyle x_i<\alpha <x_j}$. Replacing x_i by ${\displaystyle \alpha }$ and x_j by ${\displaystyle (x_i+x_j-\alpha )}$ will leave the arithmetic mean of the numbers unchanged, but will increase the geometric mean because

${\displaystyle \alpha (x_j+x_i-\alpha )-x_i{x}_j=(\alpha -x_i)(x_j-\alpha )>0}$

If the numbers are still not equal, we continue replacing numbers as above. After at most ${\displaystyle (n-1)}$ such replacement steps all the numbers will have been replaced with ${\displaystyle \alpha }$ while the geometric mean strictly increases at each step. After the last step, the geometric mean will be ${\displaystyle \sqrt[n]{\alpha \alpha \cdots \alpha }=\alpha }$, 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 ${\displaystyle x_i,x_j}$ such that ${\displaystyle 0<x_i<\beta <x_j}$. Replacing ${\displaystyle x_i}$ by ${\displaystyle \beta }$ and ${\displaystyle x_j}$ by ${\displaystyle \frac{x_i{x}_j}{\beta }}$leaves the geometric mean unchanged but strictly decreases the arithmetic mean since

${\displaystyle x_i+x_j-\beta -\frac{x_i{x}_j}{\beta }=\frac{(\beta -x_i)(x_j-\beta )}{\beta }>0}$. The proof then follows along similar lines as in the earlier replacement.

Induction proofs

Proof by induction #1

Of the non-negative real numbers x₁, . . . , x_n, the AM–GM statement is equivalent to

${\displaystyle {\alpha }^n\ge x_1{x}_2\cdots x_n}$

with equality if and only if α = x_i 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 x₁, . . . , x_n+1, . Their arithmetic mean α satisfies

${\displaystyle (n+1)\alpha =x_1+\cdots +x_n+x_{n+1}.}$

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

${\displaystyle x_n-\alpha >0\alpha -x_{n+1}>0}$

${\displaystyle ⟹(x_n-\alpha )(\alpha -x_{n+1})>0.(\ast )}$

Now define y with

${\displaystyle y:=x_n+x_{n+1}-\alpha \ge x_n-\alpha >0,}$

and consider the n numbers x₁, . . . , x_n–1, y which are all non-negative. Since

${\displaystyle (n+1)\alpha =x_1+\cdots +x_{n-1}+x_n+x_{n+1}}$

${\displaystyle n\alpha =x_1+\cdots +x_{n-1}+\underset{=y}{\underbrace{x_n+x_{n+1}-\alpha }},}$

Thus, α is also the arithmetic mean of n numbers x₁, . . . , x_n–1, y and the induction hypothesis implies

${\displaystyle {\alpha }^{n+1}={\alpha }^n\cdot \alpha \ge x_1{x}_2\cdots x_{n-1}y\cdot \alpha .(\ast \ast )}$

Due to (*) we know that

${\displaystyle (\underset{=y}{\underbrace{x_n+x_{n+1}-\alpha }})\alpha -x_n{x}_{n+1}=(x_n-\alpha )(\alpha -x_{n+1})>0,}$

hence

${\displaystyle y\alpha >x_n{x}_{n+1},(\ast \ast \ast )}$

in particular α > 0. Therefore, if at least one of the numbers x₁, . . . , 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

${\displaystyle {\alpha }^{n+1}>x_1{x}_2\cdots x_{n-1}{x}_n{x}_{n+1},}$

which completes the proof.

Proof by induction #2

First of all we shall prove that for real numbers x₁ < 1 and x₂ > 1 there follows

${\displaystyle x_1+x_2>x_1{x}_2+1.}$

Indeed, multiplying both sides of the inequality x₂ > 1 by 1 – x₁, gives

${\displaystyle x_2-x_1{x}_2>1-x_1,}$

whence the required inequality is obtained immediately.

Now, we are going to prove that for positive real numbers x₁, . . . , x_n satisfying x₁ . . . x_n = 1, there holds

${\displaystyle x_1+\cdots +x_n\ge n.}$

The equality holds only if x₁ = ... = 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 x₁, . . . , x_n, there holds x₁ . . . 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 x₁ . . . x_n = 1 we shall write in the form of (x₁ . . . x_n–2) (x_n–1 x_n) = 1. Then, the induction hypothesis implies

${\displaystyle (x_1+\cdots +x_{n-2})+(x_{n-1}{x}_n)>n-1.}$

However, taking into account the induction basis, we have

${\displaystyle \begin{array}{cc}{x}_1+\cdots +x_{n-2}+x_{n-1}+x_n& =(x_1+\cdots +x_{n-2})+(x_{n-1}+x_n)\\ & >(x_1+\cdots +x_{n-2})+x_{n-1}{x}_n+1\\ & >n,\end{array}}$

which completes the proof.

For positive real numbers a₁, . . . , a_n, let's denote

${\displaystyle x_1=\frac{a_1}{\sqrt[n]{a_1\cdots a_n}},...,x_n=\frac{a_n}{\sqrt[n]{a_1\cdots a_n}}.}$

The numbers x₁, . . . , x_n satisfy the condition x₁ . . . x_n = 1. So we have

${\displaystyle \frac{a_1}{\sqrt[n]{a_1\cdots a_n}}+\cdots +\frac{a_n}{\sqrt[n]{a_1\cdots a_n}}\ge n,}$

whence we obtain

${\displaystyle \frac{a_1+\cdots +a_n}{n}\ge \sqrt[n]{a_1\cdots a_n},}$

with the equality holding only for a₁ = ... = a_n.

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:

${\displaystyle x_1=x_2=\cdots =x_n,}$

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:

${\displaystyle \begin{array}{cc}(\frac{x_1+x_2}{2}{)}^2-x_1{x}_2& =\frac{1}{4}(x_1^2+2x_1{x}_2+x_2^2)-x_1{x}_2\\ & =\frac{1}{4}(x_1^2-2x_1{x}_2+x_2^2)\\ & =(\frac{x_1-x_2}{2}{)}^2>0,\end{array}}$

hence

${\displaystyle \frac{x_1+x_2}{2}>\sqrt{x_1{x}_2}}$

as desired.

The subcase where n = 2k

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:

${\displaystyle \begin{array}{cc}\frac{x_1+x_2+\cdots +x_{2^k}}{2^k}& =\frac{\frac{x_1+x_2+\cdots +x_{2^{k-1}}}{2^{k-1}}+\frac{x_{2^{k-1}+1}+x_{2^{k-1}+2}+\cdots +x_{2^k}}{2^{k-1}}}{2}\\ & \ge \frac{\sqrt[2^{k-1}]{x_1{x}_2\cdots x_{2^{k-1}}}+\sqrt[2^{k-1}]{x_{2^{k-1}+1}{x}_{2^{k-1}+2}\cdots x_{2^k}}}{2}\\ & \ge \sqrt{\sqrt[2^{k-1}]{x_1{x}_2\cdots x_{2^{k-1}}}\sqrt[2^{k-1}]{x_{2^{k-1}+1}{x}_{2^{k-1}+2}\cdots x_{2^k}}}\\ & =\sqrt[2^k]{x_1{x}_2\cdots x_{2^k}}\end{array}}$

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

${\displaystyle x_1=x_2=\cdots =x_{2^{k-1}}}$

and

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

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

${\displaystyle \frac{x_1+x_2+\cdots +x_{2^k}}{2^k}>\sqrt[2^k]{x_1{x}_2\cdots x_{2^k}}}$

as desired.

The subcase where n < 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, . . . , 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:

${\displaystyle x_{n+1}=x_{n+2}=\cdots =x_m=\alpha .}$

We then have:

${\displaystyle \begin{array}{cc}\alpha & =\frac{x_1+x_2+\cdots +x_n}{n}\\ & =\frac{\frac{m}{n}\left(x_1+x_2+\cdots +x_n\right)}{m}\\ & =\frac{x_1+x_2+\cdots +x_n+\frac{(m-n)}{n}\left(x_1+x_2+\cdots +x_n\right)}{m})(\because x_1+x_2+\cdots +x_n=\frac{n(x_1+x_2+\cdots +x_n)}{n})\\ & =\frac{x_1+x_2+\cdots +x_n+\left(m-n\right)\alpha }{m}\\ & =\frac{x_1+x_2+\cdots +x_n+x_{n+1}+\cdots +x_m}{m}\\ & \ge \sqrt[m]{x_1{x}_2\cdots x_n{x}_{n+1}\cdots x_m}\\ & =\sqrt[m]{x_1{x}_2\cdots x_n{\alpha }^{m-n}},\end{array}}$

so

${\displaystyle {\alpha }^m\ge x_1{x}_2\cdots x_n{\alpha }^{m-n}}$

and

${\displaystyle \alpha \ge \sqrt[n]{x_1{x}_2\cdots x_n}}$

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 x₁, . . . , x_n, x_n+1, we need to prove that

${\displaystyle \frac{x_1+\cdots +x_n+x_{n+1}}{n+1}-(x_1\cdots x_n{x}_{n+1}{)}^{\frac{1}{n+1}}\ge 0}$

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

${\displaystyle f(t)=\frac{x_1+\cdots +x_n+t}{n+1}-(x_1\cdots x_{n}t{)}^{\frac{1}{n+1}},t>0.}$

Proving the induction step is equivalent to showing that f(t) ≥ 0 for all t > 0, with f(t) = 0 only if x₁, . . . , 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

${\displaystyle f^{\prime }(t)=\frac{1}{n+1}-\frac{1}{n+1}(x_1\cdots x_n{)}^{\frac{1}{n+1}}{t}^{-\frac{n}{n+1}},t>0.}$

A critical point t₀ has to satisfy f′(t₀) = 0, which means

${\displaystyle (x_1\cdots x_n{)}^{\frac{1}{n+1}}{t}_0^{-\frac{n}{n+1}}=1.}$

After a small rearrangement we get

${\displaystyle t_0^{\frac{n}{n+1}}=(x_1\cdots x_n{)}^{\frac{1}{n+1}},}$

and finally

${\displaystyle t_0=(x_1\cdots x_n{)}^{\frac{1}{n}},}$

which is the geometric mean of x₁, . . . , 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 t₀. Next we compute the value of the function at this global minimum:

${\displaystyle \begin{array}{cc}f(t_0)& =\frac{x_1+\cdots +x_n+(x_1\cdots x_n{)}^{1/n}}{n+1}-(x_1\cdots x_n{)}^{\frac{1}{n+1}}(x_1\cdots x_n{)}^{\frac{1}{n(n+1)}}\\ & =\frac{x_1+\cdots +x_n}{n+1}+\frac{1}{n+1}(x_1\cdots x_n{)}^{\frac{1}{n}}-(x_1\cdots x_n{)}^{\frac{1}{n}}\\ & =\frac{x_1+\cdots +x_n}{n+1}-\frac{n}{n+1}(x_1\cdots x_n{)}^{\frac{1}{n}}\\ & =\frac{n}{n+1}(\frac{x_1+\cdots +x_n}{n}-(x_1\cdots x_n{)}^{\frac{1}{n}})\\ & \ge 0,\end{array}}$

where the final inequality holds due to the induction hypothesis. The hypothesis also says that we can have equality only when x₁, . . . , x_n are all equal. In this case, their geometric mean t₀ has the same value, Hence, unless x₁, . . . , 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 Rⁿ.

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 ≤ e^x–1 for all real x with equality only for x = 1.

Consider a list of non-negative real numbers x₁, x₂, . . . , 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

${\displaystyle \begin{array}{cc}\frac{x_1}{\alpha }\frac{x_2}{\alpha }\cdots \frac{x_n}{\alpha }& \le e^{\frac{x_1}{\alpha }-1}{e}^{\frac{x_2}{\alpha }-1}\cdots e^{\frac{x_n}{\alpha }-1}\\ & =\exp (\frac{x_1}{\alpha }-1+\frac{x_2}{\alpha }-1+\cdots +\frac{x_n}{\alpha }-1),(\ast )\end{array}}$

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

${\displaystyle \begin{array}{cc}\frac{x_1}{\alpha }-1+\frac{x_2}{\alpha }-1+\cdots +\frac{x_n}{\alpha }-1& =\frac{x_1+x_2+\cdots +x_n}{\alpha }-n\\ & =\frac{n\alpha }{\alpha }-n\\ & =0.\end{array}}$

Returning to (*),

${\displaystyle \frac{x_1{x}_2\cdots x_n}{{\alpha }^n}\le e^0=1,}$

which produces x₁ x₂ · · · x_n ≤ αⁿ, hence the result

${\displaystyle \sqrt[n]{x_1{x}_2\cdots x_n}\le \alpha .}$

Proof by Lagrangian multipliers

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

Because the arithmetic and geometric means are homogeneous of degree 1, without loss of generality assume that ${\displaystyle \prod _{i=1}^n{x}_i=1}$. Set ${\displaystyle G(x_1,x_2,\dots ,x_n)=\prod _{i=1}^n{x}_i}$, and ${\displaystyle F(x_1,x_2,\dots ,x_n)=\frac{1}{n}\sum _{i=1}^n{x}_i}$. The inequality will be proved (together with the equality case) if we can show that the minimum of ${\displaystyle F(x_1,x_2,...,x_n),}$ subject to the constraint ${\displaystyle G(x_1,x_2,\dots ,x_n)=1,}$ is equal to ${\displaystyle 1}$, and the minimum is only achieved when ${\displaystyle x_1=x_2=\cdots =x_n=1}$. Let us first show that the constrained minimization problem has a global minimum.

Set ${\displaystyle K=\{(x_1,x_2,\dots ,x_n):0\le x_1,x_2,\dots ,x_n\le n\}}$. Since the intersection ${\displaystyle K\cap \{G=1\}}$ is compact, the extreme value theorem guarantees that the minimum of ${\displaystyle F(x_1,x_2,...,x_n)}$ subject to the constraints ${\displaystyle G(x_1,x_2,\dots ,x_n)=1}$ and ${\displaystyle (x_1,x_2,\dots ,x_n)\in K}$ is attained at some point inside ${\displaystyle K}$. On the other hand, observe that if any of the ${\displaystyle x_i>n}$, then ${\displaystyle F(x_1,x_2,\dots ,x_n)>1}$, while ${\displaystyle F(1,1,\dots ,1)=1}$, and ${\displaystyle (1,1,\dots ,1)\in K\cap \{G=1\}}$. This means that the minimum inside ${\displaystyle K\cap \{G=1\}}$ is in fact a global minimum, since the value of ${\displaystyle F}$ at any point inside ${\displaystyle K\cap \{G=1\}}$ is certainly no smaller than the minimum, and the value of ${\displaystyle F}$ at any point ${\displaystyle (y_1,y_2,\dots ,y_n)}$ not inside ${\displaystyle K}$ is strictly bigger than the value at ${\displaystyle (1,1,\dots ,1)}$, which is no smaller than the minimum.

The method of Lagrange multipliers says that the global minimum is attained at a point ${\displaystyle (x_1,x_2,\dots ,x_n)}$ where the gradient of ${\displaystyle F(x_1,x_2,\dots ,x_n)}$ is ${\displaystyle \lambda }$ times the gradient of ${\displaystyle G(x_1,x_2,\dots ,x_n)}$, for some ${\displaystyle \lambda }$. We will show that the only point at which this happens is when ${\displaystyle x_1=x_2=\cdots =x_n=1}$ and ${\displaystyle F(x_1,x_2,...,x_n)=1.}$

Compute ${\displaystyle \frac{\mathrm{\partial }F}{\mathrm{\partial }{x}_i}=\frac{1}{n}}$ and

${\displaystyle \frac{\mathrm{\partial }G}{\mathrm{\partial }{x}_i}=\prod _{j\ne i}{x}_j=\frac{G(x_1,x_2,\dots ,x_n)}{x_i}=\frac{1}{x_i}}$

along the constraint. Setting the gradients proportional to one another therefore gives for each ${\displaystyle i}$ that ${\displaystyle \frac{1}{n}=\frac{\lambda }{x_i},}$ and so ${\displaystyle n\lambda =x_i.}$ Since the left-hand side does not depend on ${\displaystyle i}$, it follows that ${\displaystyle x_1=x_2=\cdots =x_n}$, and since ${\displaystyle G(x_1,x_2,\dots ,x_n)=1}$, it follows that ${\displaystyle x_1=x_2=\cdots =x_n=1}$ and ${\displaystyle F(x_1,x_2,\dots ,x_n)=1}$, 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 x₁, x₂, . . . , x_n and the nonnegative weights w₁, w₂, . . . , w_n be given. Set w = w₁ + w₂ + · · · + w_n. If w > 0, then the inequality

${\displaystyle \frac{w_1{x}_1+w_2{x}_2+\cdots +w_n{x}_n}{w}\ge \sqrt[w]{x_1^{w_1}{x}_2^{w_2}\cdots x_n^{w_n}}}$

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.

One stronger version of this, which also gives strengthened version of the unweighted version, is due to Aldaz. Specifically, let the nonnegative numbers x₁, x₂, . . . , x_n and the nonnegative weights w₁, w₂, . . . , w_n be given. Assume further that the sum of the weights is 1. Then

${\displaystyle \sum _{i=1}^n{w}_i{x}_i\ge \prod _{i=1}^n{x}_i^{w_i}+\sum _{i=1}^n{w}_i{\left(x_i^{\frac{1}{2}}-\sum _{k=1}^n{w}_k{x}_k^{\frac{1}{2}}\right)}^2}$.

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

${\displaystyle \begin{array}{cc}\ln (\frac{w_1{x}_1+\cdots +w_n{x}_n}{w})& >\frac{w_1}{w}\ln x_1+\cdots +\frac{w_n}{w}\ln x_n\\ & =\ln \sqrt[w]{x_1^{w_1}{x}_2^{w_2}\cdots x_n^{w_n}}.\end{array}}$

Since the natural logarithm is strictly increasing,

${\displaystyle \frac{w_1{x}_1+\cdots +w_n{x}_n}{w}>\sqrt[w]{x_1^{w_1}{x}_2^{w_2}\cdots x_n^{w_n}}.}$

Matrix arithmetic–geometric mean inequality

Most matrix generalizations of the arithmetic geometric mean inequality apply on the level of unitarily invariant norms, since, even if the matrices ${\displaystyle A}$ and ${\displaystyle B}$ are positive semi-definite, the matrix ${\displaystyle AB}$ may not be positive semi-definite and hence may not have a canonical square root. In Bhatia and Kittaneh proved that for any unitarily invariant norm ${\displaystyle |||\cdot |||}$ and positive semi-definite matrices ${\displaystyle A}$ and ${\displaystyle B}$ it is the case that

${\displaystyle |||AB|||\le \frac{1}{2}|||A^2+B^2|||}$

Later, in the same authors proved the stronger inequality that

${\displaystyle |||AB|||\le \frac{1}{4}|||(A+B{)}^2|||}$

Finally, it is known for dimension ${\displaystyle n=2}$ that the following strongest possible matrix generalization of the arithmetic-geometric mean inequality holds, and it is conjectured to hold for all ${\displaystyle n}$

${\displaystyle |||(AB{)}^{\frac{1}{2}}|||\le \frac{1}{2}|||A+B|||}$

This conjectured inequality was shown by Stephen Drury in 2012. Indeed, he proved

${\displaystyle \sqrt{{\sigma }_j(AB)}\le \frac{1}{2}{\lambda }_j(A+B),j=1,\dots ,n.}$

Finance: Link to geometric asset returns

In finance much research is concerned with accurately estimating the rate of return of an asset over multiple periods in the future. In the case of lognormal asset returns, there is an exact formula to compute the arithmetic asset return from the geometric asset return.

For simplicity, assume we are looking at yearly geometric returns r₁, r₂, ... , r_N over a time horizon of N years, i.e.

${\displaystyle r_n=\frac{V_n-V_{n-1}}{V_{n-1}},}$

where:

${\displaystyle V_n}$ = value of the asset at time ${\displaystyle n}$,

${\displaystyle V_{n-1}}$ = value of the asset at time ${\displaystyle n-1}$.

The geometric and arithmetic returns are respectively defined as

${\displaystyle g_N={\left(\prod _{n=1}^N(1+r_n)\right)}^{1/N}-1,}$

${\displaystyle a_N=\frac{1}{N}\sum _{n=1}^N{r}_n.}$

When the yearly geometric asset returns are lognormally distributed, then the following formula can be used to convert the geometric average return to the arithmetic average return:

${\displaystyle 1+g_N=\frac{1+a_N}{\sqrt{1+\frac{{\sigma }^2}{(1+a_N{)}^2}}},}$

where ${\displaystyle {\sigma }^2}$ is the variance of the observed asset returns This implicit equation for a_N can be solved exactly as follows. First, notice that by setting

${\displaystyle z=(1+a_N{)}^2,}$

we obtain a polynomial equation of degree 2:

${\displaystyle z^2-(1+g{)}^2-(1+g{)}^2{\sigma }^2=0.}$

Solving this equation for z and using the definition of z, we obtain 4 possible solutions for a_N:

${\displaystyle a_N=\pm \frac{1+g_N}{\sqrt{2}}\sqrt{1\pm \sqrt{1+\frac{4{\sigma }^2}{(1+g_N{)}^2}}}-1.}$

However, notice that

${\displaystyle \sqrt{1+\frac{4{\sigma }^2}{(1+g_N{)}^2}}\ge 1.}$

This implies that the only 2 possible solutions are (as asset returns are real numbers):

${\displaystyle a_N=\pm \frac{1+g_N}{\sqrt{2}}\sqrt{1+\sqrt{1+\frac{4{\sigma }^2}{(1+g_N{)}^2}}}-1.}$

Finally, we expect the derivative of a_N with respect to g_N to be non-negative as an increase in the geometric return should never cause a decrease in the arithmetic return. Indeed, both measure the average growth of an asset's value and therefore should move in similar directions. This leaves us with one solution to the implicit equation for a_N, namely

${\displaystyle a_N=\frac{1+g_N}{\sqrt{2}}\sqrt{1+\sqrt{1+\frac{4{\sigma }^2}{(1+g_N{)}^2}}}-1.}$

Therefore, under the assumption of lognormally distributed asset returns, the arithmetic asset return is fully determined by the geometric asset return.

Other generalizations

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

• Muirhead's inequality,
• Maclaurin's inequality,
• QM-AM-GM-HM inequalities,
• Generalized mean inequality,
• Means of complex numbers.