AM–GM inequality

Proof without wordsof theAM–GM inequality:
PR is the diameter of a circle centered on O; its radius AO is thearithmetic meanofaandb.Using thegeometric mean theorem,triangle PGR'saltitudeGQ is thegeometric mean.For any ratioa:b,AO ≥ GQ.

Visual proofthat(x+y)²≥ 4xy.Takingsquare rootsand dividing by two gives the AM–GM inequality.

Inmathematics,theinequality of arithmetic and geometric means,or more briefly theAM–GM inequality,states that thearithmetic meanof a list of non-negativereal numbersis greater than or equal to thegeometric meanof the same list; and further, that the two means are equalif and only ifevery number in the list is the same (in which case they are both that number).

The simplest non-trivial case – i.e., with more than one variable – for two non-negative numbersxandy,is the statement that

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

with equality if and only ifx=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±b)²=a²± 2ab+b²of thebinomial 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≥ 4xy,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 and then dividing both sides by2.

For a geometrical interpretation, consider arectanglewith sides of lengthxandy,hence it hasperimeter2x+ 2yandareaxy.Similarly, asquarewith all sides of length√xyhas the perimeter4√xyand the same area as the rectangle. The simplest non-trivial case of the AM–GM inequality implies for the perimeters that2x+ 2y≥ 4√xyand that only the square has the smallest perimeter amongst all rectangles of equal area.

The simplest case is implicit inEuclid's Elements,Book 5, Proposition 25.

Extensions of the AM–GM inequality are available to includeweightsorgeneralized means.

Background

Thearithmetic mean,or less precisely theaverage,of a list ofnnumbersx₁,x₂,...,x_nis the sum of the numbers divided byn:

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

Thegeometric meanis similar, except that it is only defined for a list ofnonnegativereal numbers, and uses multiplication and arootin place of addition and division:

${\sqrt[{n}]{x_{1}\cdot x_{2}\cdots x_{n}}}.$

Ifx₁,x₂,...,x_n> 0,this is equal to theexponentialof the arithmetic mean of thenatural logarithmsof the numbers:

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

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

The inequality

Restating the inequality using mathematical notation, we have that for any list ofnnonnegative real numbersx₁,x₂,...,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 ifx₁=x₂= · · · =x_n.

Geometric interpretation

In two dimensions,2x₁+ 2x₂is theperimeterof a rectangle with sides of lengthx₁andx₂.Similarly,4√x₁x₂is the perimeter of a square with the samearea,x₁x₂,as that rectangle. Thus forn= 2the 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 tondimensions. Consider ann-dimensional box with edge lengthsx₁,x₂,...,x_n. Every vertex of the box is connected tonedges of different directions, so the average length of edges incident to the vertex is(x₁+x₂+ · · · +x_n)/n. On the other hand,${\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}$is the edge length of ann-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 then-cubehas the smallest average length of edges connected to each vertex amongst alln-dimensional boxes with the same volume.

Examples

Example 1

If$a,b,c>0$,then the A.M.-G.M. tells us that

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

Example 2

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

$1+2+\dots +n\geq n{\sqrt[{n}]{n!}}$

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

and so

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

with equality at$n=1$.

Equivalently,

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

Example 3

Consider the function

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

for all positive real numbersx,yandz.Suppose we wish to find the minimal value of this function. It can be rewritten as:

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

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 forn= 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

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

Applications

An important practical application infinancial mathematicsis to computing therate of return:theannualized 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 analyzinginvestments,as the average return overstates the cumulative effect. It can also be used to prove theCauchy–Schwarz inequality.

Proofs of the AM–GM inequality

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

Proof using Jensen's inequality

Jensen's inequalitystates that the value of aconcave functionof an arithmetic mean is greater than or equal to the arithmetic mean of the function's values. Since thelogarithmfunction is concave, we have

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

Takingantilogsof the far left and far right sides, we have the AM–GM inequality.

Proof by successive replacement of elements

We have to show that

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

with equality only when all numbers are equal.

If not all numbers are equal, then there exist$x_{i},x_{j}$such that$x_{i}<\alpha <x_{j}$.Replacingx_iby$\alpha$andx_jby$(x_{i}+x_{j}-\alpha )$will leave the arithmetic mean of the numbers unchanged, but will increase the geometric mean because

$\alpha (x_{j}+x_{i}-\alpha )-x_{i}x_{j}=(\alpha -x_{i})(x_{j}-\alpha )>0$

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

$x_{i}+x_{j}-\beta -{\frac {x_{i}x_{j}}{\beta }}={\frac {(\beta -x_{i})(x_{j}-\beta )}{\beta }}>0$.The proof then follows along similar lines as in the earlier replacement.

Induction proofs

Proof by induction #1

Of the non-negative real numbersx₁,...,x_n,the AM–GM statement is equivalent to

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

with equality if and only ifα=x_ifor alli∈ {1,...,n}.

For the following proof we applymathematical inductionand only well-known rules of arithmetic.

Induction basis:Forn= 1the statement is true with equality.

Induction hypothesis:Suppose that the AM–GM statement holds for all choices ofnnon-negative real numbers.

Induction step:Considern+ 1non-negative real numbersx₁,...,x_n+1,.Their arithmetic meanαsatisfies

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

If all thex_iare equal toα,then we have equality in the AM–GM statement and we are done. In the case where some are not equal toα,there must exist one number that is greater than the arithmetic meanα,and one that is smaller thanα.Without loss of generality, we can reorder ourx_iin order to place these two particular elements at the end:x_n>αandx_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 defineywith

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

and consider thennumbersx₁,...,x_n–1,ywhich 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,αis also the arithmetic mean ofnnumbersx₁,...,x_n–1,yand 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 particularα> 0.Therefore, if at least one of the numbersx₁,...,x_n–1is 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 numbersx₁< 1andx₂> 1there follows

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

Indeed, multiplying both sides of the inequalityx₂> 1by1 –x₁,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 numbersx₁,...,x_nsatisfying x₁...x_n= 1,there holds

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

The equality holds only ifx₁=... =x_n= 1.

Induction basis:Forn= 2the statement is true because of the above property.

Induction hypothesis:Suppose that the statement is true for all natural numbers up ton– 1.

Induction step:Consider natural numbern,i.e. for positive real numbersx₁,...,x_n,there holdsx₁...x_n= 1.There exists at least onex_k< 1,so there must be at least onex_j> 1.Without loss of generality, we letk=n– 1andj=n.

Further, the equalityx₁...x_n= 1we shall write in the form of(x₁...x_n–2) (x_n–1x_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 numbersa₁,...,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 numbersx₁,...,x_nsatisfy the conditionx₁...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 fora₁=... =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 fromAugustin Louis Cauchyand can be found in hisCours d'analyse.

The case where all the terms are equal

If all the terms are equal:

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

then their sum isnx₁,so their arithmetic mean isx₁;and their product isx₁ⁿ,so their geometric mean isx₁;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 ifnotall the terms are equal, then the arithmetic mean is greater than the geometric mean. Clearly, this is only possible whenn> 1.

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

The subcase wheren= 2

Ifn= 2,then we have two terms,x₁andx₂,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 wheren= 2^k

Consider the case wheren= 2^k,wherekis a positive integer. We proceed by mathematical induction.

In the base case,k= 1,son= 2.We have already shown that the inequality holds whenn= 2,so we are done.

Now, suppose that for a givenk> 1,we have already shown that the inequality holds forn= 2^k−1,and we wish to show that it holds forn= 2^k.To do so, we apply the inequality twice for2^k-1numbers and once for2numbers 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 tox₁,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 all2^knumbers 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 wheren< 2^k

Ifnis not a natural power of2,then it is certainlylessthan some natural power of 2, since the sequence2, 4, 8,..., 2^k,...is unbounded above. Therefore, without loss of generality, letmbe some natural power of2that is greater thann.

So, if we haventerms, then let us denote their arithmetic mean byα,and expand our list of terms thus:

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

We then have:

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

so

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

and

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

as desired.

Proof by induction using basic calculus

The following proof uses mathematical induction and some basicdifferential calculus.

Induction basis:Forn= 1the statement is true with equality.

Induction hypothesis:Suppose that the AM–GM statement holds for all choices ofnnon-negative real numbers.

Induction step:In order to prove the statement forn+ 1non-negative real numbersx₁,...,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 then+ 1numbers 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 alln+ 1numbers are positive.

We consider the last numberx_n+1as 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 thatf(t) ≥ 0for allt> 0,withf(t) = 0only ifx₁,...,x_nandtare all equal. This can be done by analyzing thecritical pointsoffusing some basic calculus.

The firstderivativeoffis 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 pointt₀has to satisfyf′(t₀) = 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 ofx₁,...,x_n.This is the only critical point off.Sincef′′(t) > 0for allt> 0,the functionfisstrictly convexand has a strictglobal minimumatt₀.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 whenx₁,...,x_nare all equal. In this case, their geometric meant₀has the same value, Hence, unlessx₁,...,x_n,x_n+1are all equal, we havef(x_n+1) > 0.This completes the proof.

This technique can be used in the same manner to prove the generalized AM–GM inequality andCauchy–Schwarz inequalityin Euclidean spaceRⁿ.

Proof by Pólya using the exponential function

George Pólyaprovided a proof similar to what follows. Letf(x) = e^x–1–xfor all realx,with firstderivativef′(x) = e^x–1– 1and second derivativef′′(x) = e^x–1.Observe thatf(1) = 0,f′(1) = 0andf′′(x) > 0for all realx,hencefis strictly convex with the absolute minimum atx= 1.Hencex≤ e^x–1for all realxwith equality only forx= 1.

Consider a list of non-negative real numbersx₁,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 meanα> 0.Byn-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 ifx_i=αfor everyi∈ {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 producesx₁x₂· · ·x_n≤αⁿ,hence the result

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

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 constrained minimization problem has a global minimum.

Set$K=\{(x_{1},x_{2},\ldots,x_{n})\colon 0\leq x_{1},x_{2},\ldots,x_{n}\leq n\}$.Since the intersection$K\cap \{G=1\}$is compact, theextreme value theoremguarantees 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 ofLagrange multiplierssays 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 theweighted arithmetic meanandweighted geometric mean.Specifically, let the nonnegative numbersx₁,x₂,...,x_nand the nonnegative weightsw₁,w₂,...,w_nbe given. Setw=w₁+w₂+ · · · +w_n.Ifw> 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 thex_kwithw_k> 0are equal. Here the convention0⁰= 1is used.

If allw_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. In particular, There is a similar inequality for theweighted arithmetic meanandweighted geometric mean.Specifically, let the nonnegative numbersx₁,x₂,...,x_nand the nonnegative weightsw₁,w₂,...,w_nbe given. Assume further that the sum of the weights is 1. Then

$\sum _{i=1}^{n}w_{i}x_{i}\geq \prod _{i=1}^{n}x_{i}^{w_{i}}+\sum _{i=1}^{n}w_{i}\left(x_{i}^{\frac {1}{2}}-\sum _{i=1}^{n}w_{i}x_{i}^{\frac {1}{2}}\right)^{2}$.

Proof using Jensen's inequality

Using the finite form ofJensen's inequalityfor thenatural logarithm,we can prove the inequality between the weighted arithmetic mean and the weighted geometric mean stated above.

Since anx_kwith weightw_k= 0has no influence on the inequality, we may assume in the following that all weights are positive. If allx_kare 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 onex_kis 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 allx_kare positive.

Since the natural logarithm isstrictly concave,the finite form of Jensen's inequality and thefunctional equationsof 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 isstrictly increasing,

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

Matrix arithmetic–geometric mean inequality

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

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

Later, inthe 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 proved

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

Finance: Link to geometric asset returns

In finance much research is concerned with accurately estimating therate of returnof 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 returnsr₁,r₂,..., r_Nover a time horizon ofNyears, i.e.

$r_{n}={\frac {V_{n}-V_{n-1}}{V_{n-1}}},$

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

$g_{N}=\left(\prod _{n=1}^{N}(1+r_{n})\right)^{1/N},$

$a_{N}={\frac {1}{N}}\sum _{n=1}^{N}r_{n}.$

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

$1+g_{N}={\frac {1+a_{N}}{\sqrt {1+{\frac {\sigma ^{2}}{(1+a_{N})^{2}}}}}},$

where$\sigma ^{2}$is thevarianceof the observed asset returns This implicit equation fora_Ncan be solved exactly as follows. First, notice that by setting

$z=(1+a_{N})^{2},$

we obtain a polynomial equation of degree 2:

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

Solving this equation forzand using the definition ofz,we obtain 4 possible solutions fora_N:

$a_{N}=\pm {\frac {1+g_{N}}{\sqrt {2}}}{\sqrt {1\pm {\sqrt {1+{\frac {4\sigma ^{2}}{(1+g_{N})^{2}}}}}}}-1.$

However, notice that

${\sqrt {1+{\frac {4\sigma ^{2}}{(1+g_{N})^{2}}}}}\geq 1.$

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

$a_{N}=\pm {\frac {1+g_{N}}{\sqrt {2}}}{\sqrt {1+{\sqrt {1+{\frac {4\sigma ^{2}}{(1+g_{N})^{2}}}}}}}-1.$

Finally, we expect the derivative ofa_Nwith respect tog_Nto 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 fora_N,namely

$a_{N}={\frac {1+g_{N}}{\sqrt {2}}}{\sqrt {1+{\sqrt {1+{\frac {4\sigma ^{2}}{(1+g_{N})^{2}}}}}}}-1.$

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

Other generalizations

Geometricproof without wordsthatmax (a,b)>root mean square(RMS)orquadratic mean(QM)>arithmetic mean(AM)>geometric mean(GM)>harmonic mean(HM)>min (a,b)of two distinct positive numbersaandb

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

• Muirhead's inequality,
• Maclaurin's inequality,
• Generalized mean inequality,
• Means of complex numbers.

See also

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

External links

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