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| publisher = Springer
| title = The Mathematical Gardner
| year = 1981
}}</ref>]]
In [[mathematics]], the '''inequality of arithmetic and geometric means''', or more briefly the '''AM–GM inequality''', states that the [[arithmetic mean]] of a list of non-negative [[real number]]s is greater than or equal to the [[geometric mean]] of the same list; and further, that the two means are equal [[if and only if]] every number in the list is the same (in which case they are both that number).
The simplest non-trivial case – i.e., with more than one variable – for two non-negative numbers {{mvar|x}} and {{mvar|y}}, is the statement that
:<math>\frac{x+y}2 \ge \sqrt{xy}</math>
with equality if and only if {{math|''x'' {{=}} ''y''}}.
This case can be seen from the fact that the square of a real number is always non-negative (greater than or equal to zero) and from the elementary case {{math|(''a'' ± ''b'')<sup>2</sup> {{=}} ''a''<sup>2</sup> ± 2''ab'' + ''b''<sup>2</sup>}} of the [[binomial formula]]:
:<math>\begin{align}
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& = (x+y)^2 - 4xy.
\end{align}</math>
<span style="line-height:1.5">Hence {{math|(''x'' + ''y'')<sup>2</sup> ≥ 4''xy''}}, with equality precisely when {{math|(''x'' − ''y'')<sup>2</sup> {{=}} 0}}, i.e. {{math|''x'' {{=}} ''y''}}. The AM–GM inequality then follows from taking the positive square root of both sides and then dividing both sides by ''2''.</span>
For a geometrical interpretation, consider a [[rectangle]] with sides of length {{mvar|x}} and {{mvar|y}}, hence it has [[perimeter]] {{math|2''x'' + 2''y''}} and [[area]] {{mvar|xy}}. Similarly, a [[square]] with all sides of length {{math|{{radical|''xy''}}}} has the perimeter {{math|4{{radical|''xy''}}}} and the same area as the rectangle. The simplest non-trivial case of the AM–GM inequality implies for the perimeters that {{math|2''x'' + 2''y'' ≥ 4{{radical|''xy''}}}} and that only the square has the smallest perimeter amongst all rectangles of equal area.
The simplest case is implicit in [[Euclid's Elements]], Book 5, Proposition 25.<ref>{{Cite web |title=Euclid's Elements, Book V, Proposition 25 |url=https://aleph0.clarku.edu/~djoyce/java/elements/bookV/propV25.html }}</ref>
Extensions of the AM–GM inequality are available to include [[#Weighted_AM.E2.80.93GM_inequality|weights]] or [[generalized mean]]s.▼
▲Extensions of the AM–GM inequality are available to include [[#
{{TOC limit|4}}
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:<math>\exp \left( \frac{\ln {x_1} + \ln {x_2} + \cdots + \ln {x_n}}{n} \right).</math>
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 ==
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In two dimensions, {{math|2''x''<sub>1</sub> + 2''x''<sub>2</sub>}} is the [[perimeter]] of a rectangle with sides of length {{math|''x''<sub>1</sub>}} and {{math|''x''<sub>2</sub>}}. Similarly, {{math|4{{radical|''x''<sub>1</sub>''x''<sub>2</sub>}}}} is the perimeter of a square with the same [[area]], {{math|''x''<sub>1</sub>''x''<sub>2</sub>}}, 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|''x''<sub>1</sub>, ''x''<sub>2</sub>, . . . , ''x<sub>n</sub>''}}.
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|(''x''<sub>1</sub> + ''x''<sub>2</sub> + · · · + ''x<sub>n</sub>'')/''n''}}.
On the other hand, <math>\sqrt[n]{x_1 x_2 \cdots x_n}</math> 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|{{mvar|n}}-cube]] has the smallest
| last = Steele
| first = J. Michael
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If <math>a,b,c>0</math>, then the A.M.-G.M. tells us that
:<math>(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}</math>
===Example 2===
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with equality at <math>n=1</math>.
Equivalently,
:<math>(n+1)^n \ge 2^nn!</math>
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:<math>(x,y,z)=\biggr(t,\sqrt[3]{2}\sqrt{3}\,t,\frac{3\sqrt{3}}{2}\,t\biggr)\quad\mbox{with}\quad t>0.</math>
==Applications==
An important practical application in [[financial mathematics]] is to computing the [[rate of return]]: the [[annualized return]], computed via the geometric mean, is less than the average annual return, computed by the arithmetic mean (or equal if all returns are equal). This is important in analyzing [[investment]]s, as the average return overstates the cumulative effect. It can also be used to prove the [[Cauchy–Schwarz inequality]].
==Proofs of the AM–GM inequality==
The AM–GM inequality is also known for the variety of methods that can be used to prove it.
===Proof using Jensen's inequality===
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:<math>\alpha = \frac{x_1+x_2+\cdots+x_n}{n} \ge \sqrt[n]{x_1x_2 \cdots x_n}=\beta</math>
with equality only when all numbers are equal.
If not all numbers are equal, then there exist <math>x_i,x_j</math> such that <math>x_i<\alpha<x_j</math>. Replacing {{mvar|x<sub>i</sub>}} by <math>\alpha</math> and {{mvar|x<sub>j</sub>}} by <math>(x_i+x_j-\alpha)</math> will leave the arithmetic mean of the numbers unchanged, but will increase the geometric mean because
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:<math> x_1 + x_2 > x_1x_2+1.</math>
Indeed, multiplying both sides of the inequality {{math|''x''<sub>2</sub> > 1}} by {{math| 1 – ''x''<sub>1</sub>}}, gives
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:<math>\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}</math>
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:<math>
\frac{x_1 + x_2}{2}
as desired.
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(in which case the first arithmetic mean and first geometric mean are both equal to {{math|''x''<sub>1</sub>}}, 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<sup>''k''</sup>}} numbers are equal, it is not possible for both inequalities to be equalities, so we know that:
:<math>\frac{x_1 + x_2 + \cdots + x_{2^k}}{2^k}
as desired.
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Set <math>K = \{(x_1,x_2,\ldots,x_n) \colon 0 \leq x_1,x_2,\ldots,x_n \leq n\}</math>. Since the intersection <math>K \cap \{G = 1\}</math> is compact, the [[extreme value theorem]] guarantees that the minimum of <math>F(x_1,x_2,...,x_n)</math> subject to the constraints <math>G(x_1,x_2,\ldots,x_n) = 1</math> and <math> (x_1,x_2,\ldots,x_n) \in K </math> is attained at some point inside <math>K</math>. On the other hand, observe that if any of the <math>x_i > n</math>, then <math>F(x_1,x_2,\ldots,x_n) > 1 </math>, while <math>F(1,1,\ldots,1) = 1</math>, and <math>(1,1,\ldots,1) \in K \cap \{G = 1\} </math>. This means that the minimum inside <math>K \cap \{G = 1\}</math> is in fact a global minimum, since the value of <math>F</math> at any point inside <math>K \cap \{G = 1\}</math> is certainly no smaller than the minimum, and the value of <math>F</math> at any point <math>(y_1,y_2,\ldots, y_n)</math> not inside <math>K</math> is strictly bigger than the value at <math>(1,1,\ldots,1)</math>, which is no smaller than the minimum.
The method of [[Lagrange multipliers]] says that the global minimum is attained at a point <math>(x_1,x_2,\ldots,x_n)</math> where the gradient of <math>F(x_1,x_2,\ldots,x_n)</math> is <math>\lambda</math> times the gradient of <math>G(x_1,x_2,\ldots,x_n)</math>, for some <math>\lambda</math>. We will show that the only point at which this happens is when <math>x_1 = x_2 = \cdots = x_n = 1</math> and <math>F(x_1,x_2,...,x_n) = 1.</math>
Compute
<math>\frac{\partial F}{\partial x_i} = \frac{1}{n}</math>
and
: <math>\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}</math>
along the constraint. Setting the gradients proportional to one another therefore gives for each <math>i</math> that <math>\frac{1}{n} = \frac{\lambda}{x_i},</math> and so <math>n\lambda= x_i.</math> Since the left-hand side does not depend on <math>i</math>, it follows that <math>x_1 = x_2 = \cdots = x_n</math>, and since <math>G(x_1,x_2,\ldots, x_n) = 1</math>, it follows that <math> x_1 = x_2 = \cdots = x_n = 1</math> and <math>F(x_1,x_2,\ldots,x_n) = 1</math>, as desired.
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If all {{math|''w<sub>k</sub>'' {{=}} 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 the [[weighted arithmetic mean]] and [[weighted geometric mean]]. Specifically, let the nonnegative numbers {{math|''x''<sub>1</sub>, ''x''<sub>2</sub>, . . . , ''x<sub>n</sub>''}} and the nonnegative weights {{math|''w''<sub>1</sub>, ''w''<sub>2</sub>, . . . , ''w<sub>n</sub>''}} be given. Assume further that the sum of the weights is 1. Then
:<math>\sum_{i=1}^n w_ix_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_ix_i^{\frac{1}{2}} \right)^2 </math>.<ref>{{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 }}</ref>
====Proof using Jensen's inequality====
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| doi = 10.1016/S0024-3795(00)00048-3
| doi-access = free
}}</ref> the same authors proved the stronger inequality that
:<math>
|||AB||| \leq \frac{1}{4}|||(A+B)^2|||
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</math>
This conjectured inequality was shown by Stephen Drury in 2012. Indeed, he proved<ref>S.W. Drury, On a question of Bhatia and Kittaneh, Linear Algebra Appl. 437 (2012) 1955–1960.</ref>
:<math>\sqrt{\sigma_j(AB)}\leq \frac{1}{2}\lambda_j(A+B), \ j=1, \ldots, n.</math>
===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 {{math|''r<sub>1</sub>, r<sub>2</sub>, ... , r<sub>N</sub>''}} over a time horizon of {{mvar|N}} years, i.e.
:<math>r_n=\frac{V_n - V_{n-1}}{V_{n-1}},</math>
where:
:<math>V_n</math> = value of the asset at time <math>n</math>,
:<math>V_{n-1}</math> = value of the asset at time <math>n-1</math>.
The geometric and arithmetic returns are respectively defined as
:<math>g_N=\left(\prod_{n = 1}^N(1+r_n)\right)^{1/N},</math>
:<math>a_N=\frac1N \sum_{n = 1}^Nr_n.</math>
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:<ref>cf. Mindlin, Dimitry, On the Relationship between Arithmetic and Geometric Returns (August 14, 2011). Available at SSRN: https://ssrn.com/abstract=2083915 or http://dx.doi.org/10.2139/ssrn.2083915</ref>
:<math>1+g_N=\frac{1+a_N}{\sqrt{1+\frac{\sigma^2}{(1+a_N)^2}}},</math>
where <math>\sigma^2</math> is the [[variance]] of the observed asset returns This implicit equation for {{mvar|a<sub>N</sub>}} can be solved exactly as follows. First, notice that by setting
:<math>z=(1+a_N)^2,</math>
we obtain a polynomial equation of degree 2:
:<math>z^2 - (1+g)^2 - (1+g)^2\sigma^2 = 0.</math>
Solving this equation for {{mvar|z}} and using the definition of {{mvar|z}}, we obtain 4 possible solutions for {{mvar|a<sub>N</sub>}}:
:<math>a_N = \pm \frac{1+g_N}{\sqrt{2}}\sqrt{1 \pm \sqrt{1+\frac{4\sigma^2}{(1+g_N)^2}}}-1.</math>
However, notice that
:<math> \sqrt{1+\frac{4\sigma^2}{(1+g_N)^2}} \geq 1. </math>
This implies that the only 2 possible solutions are (as asset returns are real numbers):
:<math>a_N = \pm \frac{1+g_N}{\sqrt{2}}\sqrt{1 + \sqrt{1+\frac{4\sigma^2}{(1+g_N)^2}}}-1.</math>
Finally, we expect the derivative of {{mvar|a<sub>N</sub>}} with respect to {{mvar|g<sub>N</sub>}} 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<sub>N</sub>}}, namely
:<math>a_N = \frac{1+g_N}{\sqrt{2}}\sqrt{1 + \sqrt{1+\frac{4\sigma^2}{(1+g_N)^2}}}-1.</math>
Therefore, under the assumption of lognormally distributed asset returns, the arithmetic asset return is fully determined by the geometric asset return.
===Other generalizations===
{{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]],
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==See also==
* [[Hoffman's packing puzzle]]
* [[Ky Fan inequality]]
* [[Young's inequality for products]]
== Notes ==
{{notelist}}
{{reflist|group=note}}
==References==
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