Inequality of the arithmetic and geometric mean

In mathematics , the inequality of the arithmetic and geometric mean says that the arithmetic mean is at least as large as the geometric mean . For this inequality was already Euclid known; the first evidence for any value of was published by Colin Maclaurin in 1729 . ${\displaystyle n=2}$ ${\displaystyle n}$

Formal formulation

The inequality of the arithmetic and geometric mean reads for nonnegative numbers ${\displaystyle x_{1},x_{2},\ldots ,x_{n}}$

${\displaystyle {\sqrt[{n}]{x_{1}\cdot x_{2}\cdot \ldots \cdot x_{n}}}\leq {\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}.}$

The left side of the inequality is the geometric mean and the right side is the arithmetic mean . Equality applies if and only if applies. ${\displaystyle x_{1}=\dots =x_{n}}$

Geometric interpretation

A rectangle with the sides and has the total perimeter . A square with the same area has the perimeter . For says the inequality ${\displaystyle x_{1}}$ ${\displaystyle x_{2}}$ ${\displaystyle 2x_{1}+2x_{2}}$ ${\displaystyle 4{\sqrt {x_{1}\cdot x_{2}}}}$ ${\displaystyle n=2}$

${\displaystyle {\frac {x_{1}+x_{2}}{2}}\geq {\sqrt {x_{1}\cdot x_{2}}}}$

so that under all rectangles with the same content the scope is at least ${\displaystyle A=x_{1}\cdot x_{2}}$

${\displaystyle 2x_{1}+2x_{2}\geq 4{\sqrt {x_{1}\cdot x_{2}}}=4{\sqrt {A}}}$

where the square has this smallest circumference.

In the case , the inequality states that among all cuboids with the same volume, the cube has the smallest edge length overall. The general inequality extends this idea to dimensions . ${\displaystyle n=3}$ ${\displaystyle n}$

Inequality: ${\displaystyle {\sqrt {ab}}\leq {\frac {a+b}{2}}}$

If one plots on a straight line for the lengths and one behind the other and creates a semicircle over the ends of the line with length , the radius of that corresponds to the arithmetic mean. The geometric mean is then the length of the perpendicular of such a point on the semicircle on the line with length for which the perpendicular goes through the transition point of the lines and . The latter connection follows from the Thales theorem and the height theorem . ${\displaystyle n=2}$ ${\displaystyle a}$ ${\displaystyle b}$ ${\displaystyle a+b}$ ${\displaystyle a+b}$ ${\displaystyle a}$ ${\displaystyle b}$

proofs

In the event that a is equal to zero, the geometric mean is zero and the inequality is obviously satisfied; in the following proofs can therefore be assumed. ${\displaystyle x_{i}\!}$ ${\displaystyle x_{i}>0\!}$

Proof from those inequality

The inequality of the arithmetic and geometric mean can be proven, for example, from Jensen's inequality : the logarithm function is concave , so the following applies

${\displaystyle \ln(\lambda _{1}x_{1}+\dots +\lambda _{n}x_{n})\geq \lambda _{1}\ln x_{1}+\dots +\lambda _{n}\ln x_{n}}$

for positive with . ${\displaystyle \lambda _{i}\;}$ ${\displaystyle \sum _{i=1}^{n}\lambda _{i}=1}$

Applying the exponential function to both sides follows

${\displaystyle \lambda _{1}x_{1}+\dots +\lambda _{n}x_{n}\geq \prod _{i=1}^{n}{x_{i}}^{\lambda _{i}}}$.

For this gives exactly the inequality of the arithmetic and geometric mean. ${\displaystyle \lambda _{1}=\lambda _{2}=\cdots =\lambda _{n}=1/n}$

Proof of Polya

A proof comes from George Polya which only assumes the relationship of the exponential function . For true even ${\displaystyle \exp(x)\geq 1+x}$ ${\displaystyle x_{i}/{\bar {x}}_{\mathrm {arithm} }-1}$

${\displaystyle \exp \left(x_{i}/{\bar {x}}_{\mathrm {arithm} }-1\right)\geq x_{i}/{\bar {x}}_{\mathrm {arithm} }}$.

Multiplying these inequalities for , one obtains ${\displaystyle i=1,\dots ,n}$

${\displaystyle \exp \left(\sum _{i}x_{i}/{\bar {x}}_{\mathrm {arithm} }-n\right)\geq \prod _{i}\left(x_{i}/{\bar {x}}_{\mathrm {arithm} }\right)}$,

so

${\displaystyle 1=\exp \left(n-n\right)\geq {\bar {x}}_{\mathrm {geom} }^{n}/{\bar {x}}_{\mathrm {arithm} }^{n}}$

and thus

${\displaystyle {\bar {x}}_{\mathrm {arithm} }^{n}\geq {\bar {x}}_{\mathrm {geom} }^{n}}$.

Inductive Evidence

The proof from Jen's inequality and the polya proof are very easy to understand, but have the disadvantage that prior knowledge about the logarithm function or the exponential function is required. For studying the sequence used in defining the exponential function

${\displaystyle \exp(x)=\lim _{n\to \infty }\left(1+{x \over n}\right)^{n}}$

but the inequality of the arithmetic and geometric mean can be helpful. Methodologically, inductive proofs are therefore often more appropriate; however, these are relatively difficult for the inequality of the arithmetic and geometric mean.

Proof with forward-backward induction

An inductive proof of the inequality of the arithmetic and geometric mean can be done with a so-called "forward-backward induction". The forward step derives from the validity of the inequality for that for and obeys the scheme of ordinary complete induction. In the so-called "backward step", the inequality for the validity for is derived from the validity . ${\displaystyle n}$ ${\displaystyle 2n}$ ${\displaystyle m}$ ${\displaystyle n<m}$

Derivation

Case 2:   ${\displaystyle n=2}$ The following applies to two elements : ${\displaystyle x,y}$ ${\displaystyle {\begin{aligned}{\Bigl (}{\frac {x+y}{2}}{\Bigr )}^{2}-xy&={\frac {1}{4}}(x^{2}+2xy+y^{2})-xy\\&={\frac {1}{4}}(x^{2}-2xy+y^{2})\\&={\Bigl (}{\frac {x-y}{2}}{\Bigr )}^{2}.\end{aligned}}}$ If they are different, then is ${\displaystyle {\Bigl (}{\frac {x+y}{2}}{\Bigr )}^{2}-xy\;=\;{\Bigl (}{\frac {x-y}{2}}{\Bigr )}^{2}\;>0}$ and   ${\displaystyle {\frac {x+y}{2}}\qquad \quad \;\;>\;{\sqrt {xy}}\,.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \;\;{\mathsf {(2)}}}$ Case A:   is a power of two ${\displaystyle n\geq 1}$ This ascending ("forward") induction step is proven somewhat more generally: Does the induction assumption hold ${\displaystyle {\bar {x}}_{\mathrm {arithm} }:={\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}\;\;\geq \;\;{\sqrt[{n}]{x_{1}\cdot x_{2}\cdot \ldots \cdot x_{n}}}=:{\bar {x}}_{\mathrm {geom} }\qquad {\mathsf {(A)}}}$ for elements, then ${\displaystyle n\geq 1}$ ${\displaystyle {\bar {z}}_{\mathrm {arithm} }:={\frac {z_{1}+z_{2}+\cdots +z_{2n}}{2n}}\;\;\geq \;\;{\sqrt[{2n}]{z_{1}\cdot z_{2}\cdot \ldots \cdot z_{2n}}}=:{\bar {z}}_{\mathrm {geom} }}$ for elements. Proof: for is and for is posited. Then ${\displaystyle 2n}$ ${\displaystyle i\leq n}$ ${\displaystyle x_{i}:=z_{i}}$ ${\displaystyle i>n}$ ${\displaystyle y_{i-n}:=z_{i}}$ ${\displaystyle {\begin{aligned}{\bar {z}}_{\mathrm {arithm} }&={\frac {{\bar {x}}_{\mathrm {arithm} }}{2}}+{\frac {{\bar {y}}_{\mathrm {arithm} }}{2}}={\frac {{\bar {x}}_{\mathrm {arithm} }+{\bar {y}}_{\mathrm {arithm} }}{2}}\\&\,{\underset {\mathsf {(A)}}{\geq }}\;{\frac {{\bar {x}}_{\mathrm {geom} }+{\bar {y}}_{\mathrm {geom} }}{2}}\;{\underset {\mathsf {(2)}}{\geq }}\;{\sqrt {{\bar {x}}_{\mathrm {geom} }\,{\bar {y}}_{\mathrm {geom} }}}={\bar {z}}_{\mathrm {geom} }.\end{aligned}}}$ The equality requires and therefore the same and the same as well as taken together results in: all are equal. ${\displaystyle {\bar {z}}_{\mathrm {arithm} }={\bar {z}}_{\mathrm {geom} }}$ ${\displaystyle {\bar {x}}_{\mathrm {arithm} }={\bar {x}}_{\mathrm {geom} }}$ ${\displaystyle {\bar {y}}_{\mathrm {arithm} }={\bar {y}}_{\mathrm {geom} },}$ ${\displaystyle x_{i}={\bar {x}}_{\mathrm {geom} }}$ ${\displaystyle y_{i}={\bar {y}}_{\mathrm {geom} },}$ ${\displaystyle {\bar {x}}_{\mathrm {geom} }={\bar {y}}_{\mathrm {geom} }.}$ ${\displaystyle z_{i}}$ Case B:   is not a power of two ${\displaystyle n}$ (This part of the proof is known as the "backward" induction step.) For each there is a with . For abbreviation, use and as well . In case A the inequality for elements has already been proven, from which it follows: ${\displaystyle n\in \mathbb {N} }$ ${\displaystyle k\in \mathbb {N} }$ ${\displaystyle n<2^{k}}$ ${\displaystyle m:=2^{k}}$ ${\displaystyle z:={\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}}$ ${\displaystyle x_{n+1}:=x_{n+2}:=\cdots :=x_{m}:=z}$ ${\displaystyle m=2^{k}}$ ${\displaystyle {\frac {x_{1}+x_{2}+\cdots +x_{2^{k}}}{2^{k}}}\;\;\geq \;\;{\sqrt[{2^{k}}]{x_{1}x_{2}\cdots x_{2^{k}}}}\qquad \qquad \qquad \qquad \qquad \quad {\mathsf {(B)}}}$ Thus it follows for : ${\displaystyle n<2^{k}=m}$ ${\displaystyle {\begin{aligned}z&={\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}\\&={\frac {{\frac {m}{n}}\left(x_{1}+x_{2}+\cdots +x_{n}\right)}{m}}\\&={\frac {x_{1}+x_{2}+\cdots +x_{n}+{\frac {m-n}{n}}\left(x_{1}+x_{2}+\cdots +x_{n}\right)}{m}}\\&={\frac {x_{1}+x_{2}+\cdots +x_{n}+\left(m-n\right)z}{m}}\\&={\frac {x_{1}+x_{2}+\cdots +x_{n}\;\;+x_{n+1}+\cdots +x_{m}}{m}}\\[8.5pt]&{\underset {\mathsf {(B)}}{\geq }}\,{\sqrt[{m}]{x_{1}x_{2}\cdots x_{n}\;x_{n+1}\cdots x_{m}}}\\[5pt]&={\sqrt[{m}]{x_{1}x_{2}\cdots x_{n}z^{m-n}}}\,,\end{aligned}}}$ from what ${\displaystyle z^{m}\geq x_{1}x_{2}\cdots x_{n}z^{m-n}}$ and ${\displaystyle z^{n}\geq x_{1}x_{2}\cdots x_{n}}$ and ${\displaystyle z\geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}}$ follows. According to case A, equality only applies if all elements are equal.

This proof can already be found in Augustin Louis Cauchy .

Proof using the proposition

Another proof of the inequality of the arithmetic and geometric mean results from the proposition that for and it follows that . This proof comes from G. Ehlers. For example, the proposition can be proved with complete induction . If one looks at the product and sets , then those so defined meet the requirement of the proposition. It follows from the proposition ${\displaystyle u_{i}>0}$ ${\displaystyle \textstyle \prod _{i=1}^{n}u_{i}=1}$ ${\displaystyle \textstyle \sum _{i=1}^{n}u_{i}\geq n}$ ${\displaystyle \textstyle p:=\prod _{i=1}^{n}x_{i}}$ ${\displaystyle u_{i}:={\tfrac {x_{i}}{\sqrt[{n}]{p}}}}$ ${\displaystyle \textstyle u_{i}\!}$ ${\displaystyle \textstyle \prod _{i=1}^{n}u_{i}=1}$

${\displaystyle n\leq \sum _{i=1}^{n}u_{i}=\sum _{i=1}^{n}{\frac {x_{i}}{\sqrt[{n}]{p}}}}$,

so

${\displaystyle {\sqrt[{n}]{p}}\leq {\frac {1}{n}}\sum _{i=1}^{n}x_{i}}$.

Inserting then delivers the inequality of the arithmetic and geometric mean. ${\displaystyle \textstyle p=\prod _{i=1}^{n}x_{i}}$

Proof from the Bernoulli inequality

A direct inductive proof is possible with the help of the Bernoullian inequality : Let o. B. d. A. the maximum element of and the arithmetic mean of . Then it applies , and it follows from Bernoulli's inequality, if one “separates” the summands with the indices 1 to from the summands with the index , that ${\displaystyle x_{n+1}}$ ${\displaystyle x_{1},\dots ,x_{n},x_{n+1}}$ ${\displaystyle {\bar {x}}_{\mathrm {arithm} }}$ ${\displaystyle x_{1},\dots ,x_{n}}$ ${\displaystyle x_{n+1}-{\bar {x}}_{\mathrm {arithm} }\geq 0}$ ${\displaystyle n}$ ${\displaystyle n+1}$

${\displaystyle \left({\frac {x_{1}+\dots +x_{n+1}}{(n+1){\bar {x}}_{\mathrm {arithm} }}}\right)^{n+1}=\left(1+{\frac {x_{n+1}-{\bar {x}}_{\mathrm {arithm} }}{(n+1){\bar {x}}_{\mathrm {arithm} }}}\right)^{n+1}\geq 1+{\frac {x_{n+1}-{\bar {x}}_{\mathrm {arithm} }}{{\bar {x}}_{\mathrm {arithm} }}}={\frac {x_{n+1}}{{\bar {x}}_{\mathrm {arithm} }}}}$.

Multiplication with returns ${\displaystyle {\bar {x}}_{\mathrm {arithm} }^{n+1}}$

${\displaystyle \left({\frac {x_{1}+\dots +x_{n+1}}{n+1}}\right)^{n+1}\geq {\bar {x}}_{\mathrm {arithm} }^{n+1}{\frac {x_{n+1}}{{\bar {x}}_{\mathrm {arithm} }}}={\bar {x}}_{\mathrm {arithm} }^{n}x_{n+1}\geq x_{1}\cdots x_{n}x_{n+1}}$,

where the last inequality holds according to the induction hypothesis. Extracting the -th root ends the induction proof. ${\displaystyle (n+1)}$

This proof can be found, for example, in the textbook on analysis by H. Heuser, Part 1, Chapter 12.2.

Proof from the rearrangement inequality

A non-inductive proof of the inequality of the arithmetic and geometric mean, which manages without a logarithm function, can be carried out with the help of the rearrangement inequality . From the rearrangement inequality it follows that for positive numbers and any permutation the relationship ${\displaystyle a_{1},\dots ,a_{n}}$ ${\displaystyle a_{\sigma (1)},\dots ,a_{\sigma (n)}}$

${\displaystyle {\frac {a_{\sigma (1)}}{a_{1}}}+\cdots +{\frac {a_{\sigma (n)}}{a_{n}}}\geq n}$

must apply. If you put special

${\displaystyle a_{1}={\frac {x_{1}}{{\bar {x}}_{\mathrm {geom} }}},a_{2}={\frac {x_{1}x_{2}}{{\bar {x}}_{\mathrm {geom} }^{2}}},\dots ,a_{n}={\frac {x_{1}x_{2}\cdots x_{n}}{{\bar {x}}_{\mathrm {geom} }^{n}}}=1,}$

so it follows

${\displaystyle n\leq {\frac {a_{2}}{a_{1}}}+{\frac {a_{3}}{a_{2}}}+\dots +{\frac {a_{n}}{a_{n-1}}}+{\frac {a_{1}}{a_{n}}}={\frac {x_{2}}{{\bar {x}}_{\mathrm {geom} }}}+{\frac {x_{3}}{{\bar {x}}_{\mathrm {geom} }}}+\cdots +{\frac {x_{n}}{{\bar {x}}_{\mathrm {geom} }}}+{\frac {x_{1}}{{\bar {x}}_{\mathrm {geom} }}},}$

from which the inequality of the arithmetic and geometric mean follows immediately.

special cases

Number and its reciprocal

For , and we get: ${\displaystyle n=2}$ ${\displaystyle x_{1}=x}$ ${\displaystyle x_{2}={\tfrac {1}{x}}}$

${\displaystyle {\sqrt {x\cdot {\tfrac {1}{x}}}}\leq {\tfrac {1}{2}}\left(x+{\tfrac {1}{x}}\right)}$

${\displaystyle 1\leq {\tfrac {1}{2}}\left(x+{\tfrac {1}{x}}\right)}$

and thus

${\displaystyle 2\leq x+{\tfrac {1}{x}}}$

This statement can be proven directly: The multiplication with gives: ${\displaystyle x}$

${\displaystyle x+{\frac {1}{x}}\geq 2\Leftrightarrow x^{2}+1\geq 2x\Leftrightarrow x^{2}-2x+1\geq 0\Leftrightarrow (x-1)^{2}\geq 0,}$

which is obviously right.

Generalizations

Weighted arithmetic and geometric mean inequality

For a given positive weight tuple with and sum becomes with ${\displaystyle \mathbf {w} =(w_{1},\dots ,w_{n})}$ ${\displaystyle w_{i}>0}$ ${\displaystyle \textstyle w:={\sum _{i=1}^{n}w_{i}}}$

${\displaystyle {\bar {x}}_{\mathrm {arithm} }={\frac {\sum _{i=1}^{n}w_{i}\cdot x_{i}}{w}}}$

the weighted arithmetic mean and with

${\displaystyle {\bar {x}}_{\mathrm {geom} }={\sqrt[{w}]{\prod _{i=1}^{n}x_{i}^{w_{i}}}}}$,

denotes the weighted geometric mean. The inequality also applies to these weighted means

${\displaystyle {\bar {x}}_{\mathrm {geom} }\leq {\bar {x}}_{\mathrm {arithm} }}$.

The proof for this follows directly from the above proof with the Jensen inequality .

For , , with and , having obtained the Young's inequality ${\displaystyle n=2}$ ${\displaystyle w_{1}={\tfrac {1}{p}}}$ ${\displaystyle w_{2}={\tfrac {1}{q}}}$ ${\displaystyle w={\tfrac {1}{p}}+{\tfrac {1}{q}}=1}$ ${\displaystyle x_{1}=a^{p}}$ ${\displaystyle x_{2}=b^{q}}$ ${\displaystyle a,b\geq 0}$

${\displaystyle ab\leq {\frac {1}{p}}a^{p}+{\frac {1}{q}}b^{q}}$

Inequality of harmonic and geometric mean

If one demands really greater than zero and replaces in the inequality of the arithmetic and geometric mean with , then one obtains the inequality of the harmonic and geometric mean: ${\displaystyle x_{i}}$ ${\displaystyle x_{i}}$ ${\displaystyle 1/x_{i}}$

${\displaystyle {\frac {n}{\sum _{i=1}^{n}{\frac {1}{x_{i}}}}}\leq {\sqrt[{n}]{\prod _{i=1}^{n}{x_{i}}}}}$.

This inequality also applies to the weighted means:

${\displaystyle {\frac {w}{\sum _{i=1}^{n}{\frac {w_{i}}{x_{i}}}}}\leq {\sqrt[{w}]{\prod _{i=1}^{n}x_{i}^{w_{i}}}}}$.

Inequality of Generalized Means

The expression is called the Holder mean with exponent ${\displaystyle k}$

${\displaystyle {\bar {x}}(k)={\sqrt[{k}]{{\frac {1}{n}}\sum _{i=1}^{n}{x_{i}^{k}}}}}$.

• For one obtains the arithmetic mean, ${\displaystyle k=1\!}$
• The limit value gives the geometric mean, ${\displaystyle k\to 0}$
• For one receives the harmonic mean. ${\displaystyle k=-1}$

In general, the following applies to the generalized mean equation: ${\displaystyle -\infty \leq s\leq t\leq \infty }$

${\displaystyle {\bar {x}}(s)\leq {\bar {x}}(t)}$

This inequality can e.g. As is proved by sets and and in the Hölder's inequality with uses, or by the Jensen's inequality for convex function to the values apply. ${\displaystyle u_{i}:=x_{i}^{s},v_{i}:=1\;}$ ${\displaystyle u_{i}\;}$ ${\displaystyle v_{i}\;}$ ${\displaystyle p=t/s\;}$ ${\displaystyle f(x)=x^{t/s}\;}$ ${\displaystyle x_{i}^{s}}$

This inequality also applies to the weighted means: Let

${\displaystyle {\bar {x}}(\mathbf {w} ,k)={\sqrt[{k}]{{\frac {1}{w}}\sum _{i=1}^{n}{w_{i}x_{i}^{k}}}}}$

the weighted mean with exponent of the numbers , then applies to the inequality: ${\displaystyle \mathbf {w} }$ ${\displaystyle k}$ ${\displaystyle x_{i}}$ ${\displaystyle -\infty \leq s\leq t\leq \infty }$

${\displaystyle {\bar {x}}(\mathbf {w} ,s)\leq {\bar {x}}(\mathbf {w} ,t)}$.

This inequality can also be proven from the Hölder inequality by setting as well as , or also by applying Jensen's inequality for the convex function to the values . ${\displaystyle u_{i}:=w_{i}^{s/t}x_{i}^{s},v_{i}:=w_{i}^{1-s/t}\;}$ ${\displaystyle p=t/s\;}$ ${\displaystyle f(x)=x^{t/s}\;}$ ${\displaystyle x_{i}^{s}}$

Transferred to integrals over the measure space with a finite measure , the inequality of the generalized means takes the form ${\displaystyle (\Omega ,{\mathcal {A}},\mu )}$ ${\displaystyle \mu (\Omega )<\infty }$

${\displaystyle {\sqrt[{s}]{{\frac {1}{\mu (\Omega )}}\int _{\Omega }|f(x)|^{s}\,d\mu (x)}}\leq {\sqrt[{t}]{{\frac {1}{\mu (\Omega )}}\int _{\Omega }|f(x)|^{t}\,d\mu (x)}}}$

on; in particular it follows from this for these L ^p -spaces . ${\displaystyle L^{t}(\Omega ,{\mathcal {A}},\mu )\subseteq L^{s}(\Omega ,{\mathcal {A}},\mu )}$

See also

• Another generalization of the inequality of the arithmetic and geometric mean is the Muirhead inequality .
• The Cauchy-Schwarz inequality can be derived from the inequality of the arithmetic and geometric mean .

Individual evidence

1. ^ Paul J. Nahin : When Least is Best. Princeton University Press, Princeton NJ 2004, ISBN 0-691-07078-4 , pp. 331-333: Appendix A. The AM-GM Inequality.
2. ↑ Cauchy, Augustin-Louis. Analysis algébrique. The proof of the inequality of the arithmetic and geometric mean is on page 457 ff. There is no title in the article à la forward-backward induction.
3. ^ WD Hayes: Colloquium on linear equations. Office of Naval Research Technical Report ONRL-35-54 (1954) (PDF; 2.0 MB)

literature

• Pavel P. Korowkin : Inequality (= university books for mathematics. Small supplementary series. 4 = Mathematical student library. 9, ISSN  0076-5449 ). 6th edition. German Science Publishing House, Berlin 1970.