AM–GM inequality

The arithmetic mean, or less precisely the average, of a list of {{mvar|n}} numbers {{math|x₁, x₂, . . . , x_n}} is the sum of the numbers divided by {{mvar|n}}:

:$\backslash frac\{x\_1\; +\; x\_2\; +\; \backslash cdots\; +\; x\_n\}\{n\}.$

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:

:$\backslash sqrt[n]\{x\_1\; \backslash cdot\; x\_2\; \backslash cdots\; x\_n\}.$

If {{math|x₁, x₂, . . . , x_n > 0}}, this is equal to the exponential of the arithmetic mean of the natural logarithms of the numbers:

:$\backslash exp\; \backslash left(\; \backslash frac\{\backslash ln\; \{x\_1\}\; +\; \backslash ln\; \{x\_2\}\; +\; \backslash cdots\; +\; \backslash ln\; \{x\_n\}\}\{n\}\; \backslash right).$

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

:$\backslash frac\{x\_1\; +\; x\_2\; +\; \backslash cdots\; +\; x\_n\}\{n\}\; \backslash ge\; \backslash sqrt[n]\{x\_1\; \backslash cdot\; x\_2\; \backslash cdots\; x\_n\}\backslash ,,$

and that equality holds if and only if {{math|x₁ {{=}} x₂ {{=}} · · · {{=}} x_n}}.

In two dimensions, {{math|2x₁ + 2x₂}} is the perimeter of a rectangle with sides of length {{math|x₁}} and {{math|x₂}}. Similarly, {{math|4{{radical|x₁x₂}}}} is the perimeter of a square with the same area, {{math|x₁x₂}}, as that rectangle. Thus for {{math|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 {{mvar|n}} dimensions. Consider an {{mvar|n}}-dimensional box with edge lengths {{math|x₁, x₂, . . . , x_n}}.

Every vertex of the box is connected to {{mvar|n}} edges of different directions, so the average length of edges incident to the vertex is {{math|(x₁ + x₂ + · · · + x_n)/n}}.

On the other hand, $\backslash sqrt[n]\{x\_1\; x\_2\; \backslash cdots\; x\_n\}$ is the edge length of an {{mvar|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 Hypercube has the smallest average length of edges connected to each vertex amongst all {{mvar|n}}-dimensional boxes with the same volume.{{cite book

| last = Steele

| first = J. Michael

| title = The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities

| publisher = Cambridge University Press

| series = MAA Problem Books Series

| year = 2004

| isbn = 978-0-521-54677-5

| oclc = 54079548

}}

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

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

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

:$1+2+\backslash dots+n\; \backslash ge\; n\backslash sqrt[n]\{n!\}$

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

and so

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

with equality at $n=1$.

Equivalently,

:$(n+1)^n\; \backslash ge\; 2^nn!$

Consider the function

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

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

:$$

\begin{align}

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{align}

with

:$x\_1=\backslash frac\{x\}\{y\},\backslash qquad\; x\_2=x\_3=\backslash frac\{1\}\{2\}\; \backslash sqrt\{\backslash frac\{y\}\{z\}\},\backslash qquad\; x\_4=x\_5=x\_6=\backslash frac\{1\}\{3\}\; \backslash sqrt[3]\{\backslash frac\{z\}\{x\}\}.$

Applying the AM–GM inequality for {{math|n {{=}} 6}}, we get

:$$

\begin{align}

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{align}

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

:$f(x,y,z)\; =\; 2^\{2/3\}\; \backslash cdot\; 3^\{1/2\}\; \backslash quad\; \backslash mbox\{when\}\; \backslash quad\; \backslash frac\{x\}\{y\}\; =\; \backslash frac\{1\}\{2\}\; \backslash sqrt\{\backslash frac\{y\}\{z\}\}\; =\; \backslash frac\{1\}\{3\}\; \backslash sqrt[3]\{\backslash frac\{z\}\{x\}\}.$

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

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

The AM-GM equality can be used to prove the Cauchy–Schwarz inequality.{{Cn|date=November 2024}}

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.

The Motzkin polynomial $$x^4y^2+x^2y^4-3x^2y^2+1$$ is a nonnegative polynomial which is not a sum of square polynomials. It can be proven nonnegative using the AM-GM inequality with $x\_1\; =\; x^4\; y^2$, $x\_2\; =\; x^2\; y^4$, and $x\_3\; =\; 1$,{{cite book |last=Motzkin |first=T. S. |title=Inequalities (Proc. Sympos. Wright-Patterson Air Force Base, Ohio, 1965) |publisher=Academic Press |year=1967 |location=New york |pages=205–224 |contribution=The arithmetic-geometric inequality |mr=0223521}} that is, $$\{\backslash sqrt[3]\{(x^4\; y^2)\; \backslash cdot\; (x^2\; y^4)\; \backslash cdot\; (1)\}\}\; \backslash le\; \{\{(x^4\; y^2)+(x^2\; y^4)+(1)\}\backslash over\{3\}\}.$$ Simplifying and multiplying both sides by 3 gives $$\{3\; x^2\; y^2\}\; \backslash le\; \{x^4\; y^2\; +\; x^2\; y^4\; +\; 1\},$$ so

Aaron Potechin, Sum of Squares seminar, University of Chicago,

"[https://people.cs.uchicago.edu/~potechin/SOSseminar2017/SOSProofsandMotzkin.pdf Lecture 5: SOS Proofs and the Motzkin Polynomial]", slide 25 $$\{0\; \backslash le\; x^4\; y^2\; +\; x^2\; y^4\; -\; 3\; x^2\; y^2\; +\; 1\}.$$

The AM–GM inequality can be proven in many ways.

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

:$\backslash log\; \backslash left(\backslash frac\; \{\; \backslash sum\; x\_i\}\{n\}\; \backslash right)\; \backslash geq\; \backslash frac\{1\}\{n\}\; \backslash sum\; \backslash log\; x\_i\; =\; \backslash frac\{1\}\{n\}\; \backslash log\; \backslash left(\; \backslash prod\; x\_i\backslash right)\; =\; \backslash log\; \backslash left(\; \backslash left(\; \backslash prod\; x\_i\; \backslash right)\; ^\{1/n\}\; \backslash right).$

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

We have to show that

:$\backslash alpha\; =\; \backslash frac\{x\_1+x\_2+\backslash cdots+x\_n\}\{n\}\; \backslash ge\; \backslash sqrt[n]\{x\_1x\_2\; \backslash cdots\; x\_n\}=\backslash beta$

with equality only when all numbers are equal.

If not all numbers are equal, then there exist $x\_i,x\_j$ such that _i}} by $\backslash alpha$ and {{mvar|x_j}} by $(x\_i+x\_j-\backslash alpha)$ will leave the arithmetic mean of the numbers unchanged, but will increase the geometric mean because

:$\backslash alpha(x\_j+x\_i-\backslash alpha)-x\_ix\_j=(\backslash alpha-x\_i)(x\_j-\backslash alpha)>0$

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 $\backslash alpha$ while the geometric mean strictly increases at each step. After the last step, the geometric mean will be $\backslash sqrt[n]\{\backslash alpha\backslash alpha\; \backslash cdots\; \backslash alpha\}=\backslash alpha$, 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

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

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

:$\backslash alpha^n\backslash ge\; x\_1\; x\_2\; \backslash cdots\; x\_n$

with equality if and only if {{math|α {{=}} x_i}} for all {{math|i ∈ {1, . . . , n}}}.

For the following proof we apply mathematical induction and only well-known rules of arithmetic.

Induction basis: For {{math|n {{=}} 1}} the statement is true with equality.

Induction hypothesis: Suppose that the AM–GM statement holds for all choices of {{mvar|n}} non-negative real numbers.

Induction step: Consider {{math|n + 1}} non-negative real numbers {{math|x₁, . . . , x_n+1}}, . Their arithmetic mean {{mvar|α}} satisfies

:$(n+1)\backslash alpha=\backslash \; x\_1\; +\; \backslash cdots\; +\; x\_n\; +\; x\_\{n+1\}.$

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

:$x\_n\; -\; \backslash alpha\; >\; 0\backslash qquad\; \backslash alpha-x\_\{n+1\}>0$

:$\backslash implies\; (x\_n-\backslash alpha)(\backslash alpha-x\_\{n+1\})>0\backslash ,.\backslash qquad(*)$

Now define {{math|y}} with

:$y:=x\_n+x\_\{n+1\}-\backslash alpha\backslash ge\; x\_n-\backslash alpha>0\backslash ,,$

and consider the {{mvar|n}} numbers {{math|x₁, . . . , x_n–1, y}} which are all non-negative. Since

:$(n+1)\backslash alpha=x\_1\; +\; \backslash cdots\; +\; x\_\{n-1\}\; +\; x\_n\; +\; x\_\{n+1\}$

:$n\backslash alpha=x\_1\; +\; \backslash cdots\; +\; x\_\{n-1\}\; +\; \backslash underbrace\{x\_n+x\_\{n+1\}-\backslash alpha\}\_\{=\backslash ,y\},$

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

:$\backslash alpha^\{n+1\}=\backslash alpha^n\backslash cdot\backslash alpha\backslash ge\; x\_1x\_2\; \backslash cdots\; x\_\{n-1\}\; y\backslash cdot\backslash alpha.\backslash qquad(**)$

Due to (*) we know that

:$(\backslash underbrace\{x\_n+x\_\{n+1\}-\backslash alpha\}\_\{=\backslash ,y\})\backslash alpha-x\_nx\_\{n+1\}=(x\_n-\backslash alpha)(\backslash alpha-x\_\{n+1\})>0,$

hence

:$y\backslash alpha>x\_nx\_\{n+1\}\backslash ,,\backslash qquad(\{*\}\{*\}\{*\})$

in particular {{math|α > 0}}. Therefore, if at least one of the numbers {{math|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

:$\backslash alpha^\{n+1\}>x\_1x\_2\; \backslash cdots\; x\_\{n-1\}\; x\_nx\_\{n+1\}\backslash ,,$

which completes the proof.

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

:$x\_1\; +\; x\_2\; >\; x\_1x\_2+1.$

Indeed, multiplying both sides of the inequality {{math|x₂ > 1}} by {{math| 1 – x₁}}, gives

:$x\_2\; -\; x\_1x\_2\; >\; 1\; -\; x\_1,$

whence the required inequality is obtained immediately.

Now, we are going to prove that for positive real numbers {{math|x₁, . . . , x_n}} satisfying

{{math|x₁ . . . x_n {{=}} 1}}, there holds

:$x\_1\; +\; \backslash cdots\; +\; x\_n\; \backslash ge\; n.$

The equality holds only if {{math|x₁ {{=}} ... {{=}} x_n {{=}} 1}}.

Induction basis: For {{math|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 {{math|n – 1}}.

Induction step: Consider natural number {{math|n}}, i.e. for positive real numbers {{math|x₁, . . . , x_n}}, there holds {{math|x₁ . . . x_n {{=}} 1}}. There exists at least one {{math|x_k < 1}}, so there must be at least one {{math|x_j > 1}}. Without loss of generality, we let {{math|k {{=}}n – 1}} and {{math|j {{=}} n}}.

Further, the equality {{math|x₁ . . . x_n {{=}} 1}} we shall write in the form of {{math|(x₁ . . . x_n–2) (x_n–1 x_n) {{=}} 1}}. Then, the induction hypothesis implies

:$(x\_1\; +\; \backslash cdots\; +\; x\_\{n-2\})\; +\; (x\_\{n-1\}\; x\_n\; )\; >\; n\; -\; 1.$

However, taking into account the induction basis, we have

:$\backslash begin\{align\}$

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{align}

which completes the proof.

For positive real numbers {{math|a₁, . . . , a_n}}, let's denote

:$x\_1\; =\; \backslash frac\{a\_1\}\{\backslash sqrt[n]\{a\_1\backslash cdots\; a\_n\}\},\; .\; .\; .,\; x\_n\; =\; \backslash frac\{a\_n\}\{\backslash sqrt[n]\{a\_1\backslash cdots\; a\_n\}\}.$

The numbers {{math|x₁, . . . , x_n}} satisfy the condition {{math|x₁ . . . x_n {{=}} 1}}. So we have

:$\backslash frac\{a\_1\}\{\backslash sqrt[n]\{a\_1\backslash cdots\; a\_n\}\}\; +\; \backslash cdots\; +\; \backslash frac\{a\_n\}\{\backslash sqrt[n]\{a\_1\backslash cdots\; a\_n\}\}\; \backslash ge\; n,$

whence we obtain

:$\backslash frac\{a\_1\; +\; \backslash cdots\; +\; a\_n\}n\; \backslash ge\; \backslash sqrt[n]\{a\_1\backslash cdots\; a\_n\},$

with the equality holding only for {{math|a₁ {{=}} ... {{=}} a_n}}.

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.{{cite book |last1=Cauchy |first1=Augustin-Louis |title=Cours d'analyse de l’École royale polytechnique; I.re Partie. Analyse algébrique |author-link=Augustin Louis Cauchy |date=1821 |location=Paris |pages=457-459 |url=https://archive.org/details/coursdanalysede00caucgoog/page/458/mode/2up |chapter=Note II, Theorem 17 |lang=fr}}

If all the terms are equal:

:$x\_1\; =\; x\_2\; =\; \backslash cdots\; =\; x\_n,$

then their sum is {{math|nx₁}}, so their arithmetic mean is {{math|x₁}}; and their product is {{math|x₁ⁿ}}, so their geometric mean is {{math|x₁}}; therefore, the arithmetic mean and geometric mean are equal, as desired.

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 {{math|n > 1}}.

This case is significantly more complex, and we divide it into subcases.

If {{math|n {{=}} 2}}, then we have two terms, {{math|x₁}} and {{math|x₂}}, and since (by our assumption) not all terms are equal, we have:

:$\backslash begin\{align\}$

\Bigl(\frac{x_1+x_2}{2}\Bigr)^2-x_1x_2

&=\frac14(x_1^2+2x_1x_2+x_2^2)-x_1x_2\\

&=\frac14(x_1^2-2x_1x_2+x_2^2)\\

&=\Bigl(\frac{x_1-x_2}{2}\Bigr)^2>0,

\end{align}

hence

:$$

\frac{x_1 + x_2}{2} > \sqrt{x_1 x_2}

as desired.

Consider the case where {{math|n {{=}} 2^k}}, where {{mvar|k}} is a positive integer. We proceed by mathematical induction.

In the base case, {{math|k {{=}} 1}}, so {{math|n {{=}} 2}}. We have already shown that the inequality holds when {{math|n {{=}} 2}}, so we are done.

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

:$$

\begin{align}

\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]

& \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} \\[7pt]

& \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}}} \\[7pt]

& = \sqrt[2^k]{x_1 x_2 \cdots x_{2^k}}

\end{align}

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

:$x\_1\; =\; x\_2\; =\; \backslash cdots\; =\; x\_\{2^\{k-1\}\}$

and

:$x\_\{2^\{k-1\}+1\}\; =\; x\_\{2^\{k-1\}+2\}\; =\; \backslash cdots\; =\; x\_\{2^k\}$

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

:$\backslash frac\{x\_1\; +\; x\_2\; +\; \backslash cdots\; +\; x\_\{2^k\}\}\{2^k\}\; >\; \backslash sqrt[2^k]\{x\_1\; x\_2\; \backslash cdots\; x\_\{2^k\}\}$

as desired.

If {{mvar|n}} is not a natural power of {{math|2}}, then it is certainly less than some natural power of 2, since the sequence {{math|2, 4, 8, . . . , 2^k, . . .}} is unbounded above. Therefore, without loss of generality, let {{mvar|m}} be some natural power of {{math|2}} that is greater than {{mvar|n}}.

So, if we have {{mvar|n}} terms, then let us denote their arithmetic mean by {{mvar|α}}, and expand our list of terms thus:

:$x\_\{n+1\}\; =\; x\_\{n+2\}\; =\; \backslash cdots\; =\; x\_m\; =\; \backslash alpha.$

We then have:

: $\backslash begin\{align\}$

\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})

(\because x_1 + x_2 + \cdots + x_n = \frac{{n} (x_1 + x_2 + \cdots + x_n)}{{n}}) \\[3pt]

& = \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]

& \ge \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{align}

so

:$\backslash alpha^m\; \backslash ge\; x\_1\; x\_2\; \backslash cdots\; x\_n\; \backslash alpha^\{m-n\}$

and

:$\backslash alpha\; \backslash ge\; \backslash sqrt[n]\{x\_1\; x\_2\; \backslash cdots\; x\_n\}$

as desired.

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

Induction basis: For {{math|n {{=}} 1}} the statement is true with equality.

Induction hypothesis: Suppose that the AM–GM statement holds for all choices of {{mvar|n}} non-negative real numbers.

Induction step: In order to prove the statement for {{math|n + 1}} non-negative real numbers {{math|x₁, . . . , x_n, x_n+1}}, we need to prove that

:$\backslash frac\{x\_1\; +\; \backslash cdots\; +\; x\_n\; +\; x\_\{n+1\}\}\{n+1\}\; -\; (\{x\_1\; \backslash cdots\; x\_n\; x\_\{n+1\}\})^\{\backslash frac\{1\}\{n+1\}\}\backslash ge0$

with equality only if all the {{math|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 {{math|n + 1}} numbers are positive.

We consider the last number {{math|x_n+1}} as a variable and define the function

:$f(t)=\backslash frac\{x\_1\; +\; \backslash cdots\; +\; x\_n\; +\; t\}\{n+1\}\; -\; (\{x\_1\; \backslash cdots\; x\_n\; t\})^\{\backslash frac\{1\}\{n+1\}\},\backslash qquad\; t>0.$

Proving the induction step is equivalent to showing that {{math|f(t) ≥ 0}} for all {{math|t > 0}}, with {{math|f(t) {{=}} 0}} only if {{math|x₁, . . . , x_n}} and {{mvar|t}} are all equal. This can be done by analyzing the critical points of {{mvar|f}} using some basic calculus.

The first derivative of {{mvar|f}} is given by

:$f\text{'}(t)=\backslash frac\{1\}\{n+1\}-\backslash frac\{1\}\{n+1\}(\{x\_1\; \backslash cdots\; x\_n\})^\{\backslash frac\{1\}\{n+1\}\}t^\{-\backslash frac\{n\}\{n+1\}\},\backslash qquad\; t>0.$

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

:$(\{x\_1\; \backslash cdots\; x\_n\})^\{\backslash frac\{1\}\{n+1\}\}t\_0^\{-\backslash frac\{n\}\{n+1\}\}=1.$

After a small rearrangement we get

:$t\_0^\{\backslash frac\{n\}\{n+1\}\}=(\{x\_1\; \backslash cdots\; x\_n\})^\{\backslash frac\{1\}\{n+1\}\},$

and finally

:$t\_0=(\{x\_1\; \backslash cdots\; x\_n\})^\{\backslash frac\{1\}n\},$

which is the geometric mean of {{math|x₁, . . . , x_n}}. This is the only critical point of {{mvar|f}}. Since {{math|f′′(t) > 0}} for all {{math|t > 0}}, the function {{mvar|f}} is strictly convex and has a strict global minimum at {{math|t₀}}. Next we compute the value of the function at this global minimum:

:$$

\begin{align}

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)

\\ &\ge0,

\end{align}

where the final inequality holds due to the induction hypothesis. The hypothesis also says that we can have equality only when {{math|x₁, . . . , x_n}} are all equal. In this case, their geometric mean  {{math|t₀}} has the same value, Hence, unless {{math|x₁, . . . , x_n, x_n+1}} are all equal, we have {{math|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 {{math|Rⁿ}}.

George Pólya provided a proof similar to what follows. Let {{math|f(x) {{=}} e^x–1 – x}} for all real {{mvar|x}}, with first derivative {{math|f′(x) {{=}} e^x–1 – 1}} and second derivative {{math|f′′(x) {{=}} e^x–1}}. Observe that {{math|f(1) {{=}} 0}}, {{math|f′(1) {{=}} 0}} and {{math|f′′(x) > 0}} for all real {{mvar|x}}, hence {{mvar|f}} is strictly convex with the absolute minimum at {{math|x {{=}} 1}}. Hence {{math|x ≤ e^x–1}} for all real {{mvar|x}} with equality only for {{math|x {{=}} 1}}.

Consider a list of non-negative real numbers {{math|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 {{math|α > 0}}. By {{mvar|n}}-fold application of the above inequality, we obtain that

:

& = \exp \Bigl( \frac{x_1}{\alpha} - 1 + \frac{x_2}{\alpha} - 1 + \cdots + \frac{x_n}{\alpha} - 1 \Bigr), \qquad (*)

\end{align}

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

:$\backslash begin\{align\}$

\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{align}

Returning to {{math|(*)}},

:$\backslash frac\{x\_1\; x\_2\; \backslash cdots\; x\_n\}\{\backslash alpha^n\}\; \backslash le\; e^0\; =\; 1,$

which produces {{math|x₁ x₂ · · · x_n ≤ αⁿ}}, hence the result{{cite book

| last1 = Arnold

| first1 = Denise

| last2 = Arnold

| first2 = Graham

| title = Four unit mathematics

| publisher = Hodder Arnold H&S

| year = 1993

| isbn = 978-0-340-54335-1

| oclc = 38328013

| page = 242

}}

:$\backslash sqrt[n]\{x\_1\; x\_2\; \backslash cdots\; x\_n\}\; \backslash le\; \backslash alpha.$

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 $\backslash prod\_\{i=1\}^n\; x\_i\; =\; 1$. Set $G(x\_1,x\_2,\backslash ldots,x\_n)=\backslash prod\_\{i=1\}^n\; x\_i$, and $F(x\_1,x\_2,\backslash ldots,x\_n)\; =\; \backslash frac\{1\}\{n\}\backslash sum\_\{i=1\}^n\; x\_i$. 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 $G(x\_1,x\_2,\backslash ldots,x\_n)\; =\; 1,$ is equal to $1$, and the minimum is only achieved when $x\_1\; =\; x\_2\; =\; \backslash cdots\; =\; x\_n\; =\; 1$. Let us first show that the constrained minimization problem has a global minimum.

Set $K\; =\; \backslash \{(x\_1,x\_2,\backslash ldots,x\_n)\; \backslash colon\; 0\; \backslash leq\; x\_1,x\_2,\backslash ldots,x\_n\; \backslash leq\; n\backslash \}$. Since the intersection $K\; \backslash cap\; \backslash \{G\; =\; 1\backslash \}$ is compact, the extreme value theorem guarantees that the minimum of $F(x\_1,x\_2,...,x\_n)$ subject to the constraints $G(x\_1,x\_2,\backslash ldots,x\_n)\; =\; 1$ and $(x\_1,x\_2,\backslash ldots,x\_n)\; \backslash in\; K$ is attained at some point inside $K$. On the other hand, observe that if any of the $x\_i\; >\; n$, then $F(x\_1,x\_2,\backslash ldots,x\_n)\; >\; 1$, while $F(1,1,\backslash ldots,1)\; =\; 1$, and $(1,1,\backslash ldots,1)\; \backslash in\; K\; \backslash cap\; \backslash \{G\; =\; 1\backslash \}$. This means that the minimum inside $K\; \backslash cap\; \backslash \{G\; =\; 1\backslash \}$ is in fact a global minimum, since the value of $F$ at any point inside $K\; \backslash cap\; \backslash \{G\; =\; 1\backslash \}$ is certainly no smaller than the minimum, and the value of $F$ at any point $(y\_1,y\_2,\backslash ldots,\; y\_n)$ not inside $K$ is strictly bigger than the value at $(1,1,\backslash ldots,1)$, which is no smaller than the minimum.

The method of Lagrange multipliers says that the global minimum is attained at a point $(x\_1,x\_2,\backslash ldots,x\_n)$ where the gradient of $F(x\_1,x\_2,\backslash ldots,x\_n)$ is $\backslash lambda$ times the gradient of $G(x\_1,x\_2,\backslash ldots,x\_n)$, for some $\backslash lambda$. We will show that the only point at which this happens is when $x\_1\; =\; x\_2\; =\; \backslash cdots\; =\; x\_n\; =\; 1$ and $F(x\_1,x\_2,...,x\_n)\; =\; 1.$

Compute

$\backslash frac\{\backslash partial\; F\}\{\backslash partial\; x\_i\}\; =\; \backslash frac\{1\}\{n\}$

and

: $\backslash frac\{\backslash partial\; G\}\{\backslash partial\; x\_i\}\; =\; \backslash prod\_\{j\; \backslash neq\; i\}x\_j\; =\; \backslash frac\{G(x\_1,x\_2,\backslash ldots,x\_n)\}\{x\_i\}\; =\; \backslash frac\{1\}\{x\_i\}$

along the constraint. Setting the gradients proportional to one another therefore gives for each $i$ that $\backslash frac\{1\}\{n\}\; =\; \backslash frac\{\backslash lambda\}\{x\_i\},$ and so $n\backslash lambda=\; x\_i.$ Since the left-hand side does not depend on $i$, it follows that $x\_1\; =\; x\_2\; =\; \backslash cdots\; =\; x\_n$, and since $G(x\_1,x\_2,\backslash ldots,\; x\_n)\; =\; 1$, it follows that $x\_1\; =\; x\_2\; =\; \backslash cdots\; =\; x\_n\; =\; 1$ and $F(x\_1,x\_2,\backslash ldots,x\_n)\; =\; 1$, as desired.

There is a similar inequality for the weighted arithmetic mean and weighted geometric mean. Specifically, let the nonnegative numbers {{math|x₁, x₂, . . . , x_n}} and the nonnegative weights {{math|w₁, w₂, . . . , w_n}} be given. Set {{math|w {{=}} w₁ + w₂ + · · · + w_n}}. If {{math|w > 0}}, then the inequality

: $\backslash frac\{w\_1\; x\_1\; +\; w\_2\; x\_2\; +\; \backslash cdots\; +\; w\_n\; x\_n\}\{w\}\; \backslash ge\; \backslash sqrt[w]\{x\_1^\{w\_1\}\; x\_2^\{w\_2\}\; \backslash cdots\; x\_n^\{w\_n\}\}$

holds with equality if and only if all the {{mvar|x_k}} with {{math|w_k > 0}} are equal. Here the convention {{math|0⁰ {{=}} 1}} is used.

If all {{math|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 {{math|x₁, x₂, . . . , x_n}} and the nonnegative weights {{math|w₁, w₂, . . . , w_n}} be given. Assume further that the sum of the weights is 1. Then

:$\backslash sum\_\{i=1\}^n\; w\_ix\_i\; \backslash geq\; \backslash prod\_\{i=1\}^n\; x\_i^\{w\_i\}\; +\; \backslash sum\_\{i=1\}^n\; w\_i\backslash left(x\_i^\{\backslash frac\{1\}\{2\}\}\; -\backslash sum\_\{i=1\}^n\; w\_ix\_i^\{\backslash frac\{1\}\{2\}\}\; \backslash right)^2$.{{cite journal |last1=Aldaz |first1=J.M. |title=Self-Improvement of the Inequality Between Arithmetic and Geometric Means |journal=Journal of Mathematical Inequalities |date=2009 |volume=3 |issue=2 |pages=213–216 |doi=10.7153/jmi-03-21 |url=http://jmi.ele-math.com/03-21/Self-improvement-of-the-inequality-between-arithmetic-and-geometric-means |access-date=11 January 2023|doi-access=free }}

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 {{mvar|x_k}} with weight {{math|w_k {{=}} 0}} has no influence on the inequality, we may assume in the following that all weights are positive. If all {{mvar|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 {{mvar|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 {{mvar|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

:$\backslash begin\{align\}$

\ln\Bigl(\frac{w_1x_1+\cdots+w_nx_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{align}

Since the natural logarithm is strictly increasing,

:$$

\frac{w_1x_1+\cdots+w_nx_n}w

>\sqrt[w]{x_1^{w_1} x_2^{w_2} \cdots x_n^{w_n}}.

Most matrix generalizations of the arithmetic geometric mean inequality apply on the level of unitarily invariant norms, since, even if the matrices $A$ and $B$ are positive semi-definite, the matrix $A\; B$ may not be positive semi-definite and hence may not have a canonical square root. In {{cite journal

| last1 = Bhatia

| first1 = Rajendra

| last2= Kittaneh

| first2= Fuad

| title = On the singular values of a product of operators

| journal = SIAM Journal on Matrix Analysis and Applications

| year = 1990

|volume = 11|issue = 2|pages = 272–277| doi = 10.1137/0611018

}} Bhatia and Kittaneh proved that for any unitarily invariant norm $|||\backslash cdot|||$ and positive semi-definite matrices $A$ and $B$ it is the case that

:$$

|||AB|||\leq \frac{1}{2}|||A^2 + B^2|||

Later, in {{cite journal

| last1 = Bhatia

| first1 = Rajendra

| last2= Kittaneh

| first2= Fuad

| title = Notes on matrix arithmetic-geometric mean inequalities

| journal = Linear Algebra and Its Applications

| year = 2000

| volume = 308

| issue = 1–3

| pages = 203–211

| doi = 10.1016/S0024-3795(00)00048-3

| doi-access = free

}} the same authors proved the stronger inequality that

:$$

|||AB||| \leq \frac{1}{4}|||(A+B)^2|||

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

:$$

|||(AB)^{\frac{1}{2}}|||\leq \frac{1}{2}|||A+B|||

This conjectured inequality was shown by Stephen Drury in 2012. Indeed, he provedS.W. Drury, On a question of Bhatia and Kittaneh, Linear Algebra Appl. 437 (2012) 1955–1960.

:$\backslash sqrt\{\backslash sigma\_j(AB)\}\backslash leq\; \backslash frac\{1\}\{2\}\backslash lambda\_j(A+B),\; \backslash \; j=1,\; \backslash ldots,\; n.$

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 {{math|r₁, r₂, ... , r_N}} over a time horizon of {{mvar|N}} years, i.e.

:$r\_n=\backslash frac\{V\_n\; -\; V\_\{n-1\}\}\{V\_\{n-1\}\},$

where:

:$V\_n$ = value of the asset at time $n$,

:$V\_\{n-1\}$ = value of the asset at time $n-1$.

The geometric and arithmetic returns are respectively defined as

:$g\_N=\backslash left(\backslash prod\_\{n\; =\; 1\}^N(1+r\_n)\backslash right)^\{1/N\},$

:$a\_N=\backslash frac1N\; \backslash sum\_\{n\; =\; 1\}^Nr\_n.$

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:{{Cite journal |last=Mindlin |first=Dimitry |date=2011 |title=On the Relationship between Arithmetic and Geometric Returns |url=http://www.ssrn.com/abstract=2083915 |journal=SSRN Electronic Journal |language=en |doi=10.2139/ssrn.2083915 |issn=1556-5068|url-access=subscription }}

:$1+g\_N=\backslash frac\{1+a\_N\}\{\backslash sqrt\{1+\backslash frac\{\backslash sigma^2\}\{(1+a\_N)^2\}\}\},$

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

:$z=(1+a\_N)^2,$

we obtain a polynomial equation of degree 2:

:$z^2\; -\; (1+g)^2\; -\; (1+g)^2\backslash sigma^2\; =\; 0.$

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

:$a\_N\; =\; \backslash pm\; \backslash frac\{1+g\_N\}\{\backslash sqrt\{2\}\}\backslash sqrt\{1\; \backslash pm\; \backslash sqrt\{1+\backslash frac\{4\backslash sigma^2\}\{(1+g\_N)^2\}\}\}-1.$

However, notice that

:$\backslash sqrt\{1+\backslash frac\{4\backslash sigma^2\}\{(1+g\_N)^2\}\}\; \backslash geq\; 1.$

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

:$a\_N\; =\; \backslash pm\; \backslash frac\{1+g\_N\}\{\backslash sqrt\{2\}\}\backslash sqrt\{1\; +\; \backslash sqrt\{1+\backslash frac\{4\backslash sigma^2\}\{(1+g\_N)^2\}\}\}-1.$

Finally, we expect the derivative of {{mvar|a_N}} with respect to {{mvar|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 {{mvar|a_N}}, namely

:$a\_N\; =\; \backslash frac\{1+g\_N\}\{\backslash sqrt\{2\}\}\backslash sqrt\{1\; +\; \backslash sqrt\{1+\backslash frac\{4\backslash sigma^2\}\{(1+g\_N)^2\}\}\}-1.$

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

{{QM_AM_GM_HM_inequality_visual_proof.svg}}

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.cf. Iordanescu, R.; Nichita, F.F.; Pasarescu, O. Unification Theories: Means and Generalized Euler Formulas. Axioms 2020, 9, 144.

• Hoffman's packing puzzle
• Ky Fan inequality
• Young's inequality for products

{{notelist}}

{{reflist|group=note}}

{{Reflist}}

• {{cite web|title=Introduction to Inequalities|url=http://www.mediafire.com/file/1mw1tkgozzu |author=Arthur Lohwater|year=1982|publisher=Online e-book in PDF format}}

{{DEFAULTSORT:AM-GM inequality}}

Category:Inequalities (mathematics)

Category:Means

Category:Articles containing proofs