- Inequality of arithmetic and geometric means
In

mathematics , the**inequality of arithmetic and geometric means**, or more briefly the**AM-GM inequality**, states that thearithmetic mean of a list of non-negativereal number s is greater than or equal to thegeometric mean of the same list; and further, that the two means are equalif and only if every number in the list is the same.**Background**The "arithmetic mean", or less precisely the "average", of a list of "n" numbers "x"

_{1}, "x"_{2}, . . ., "x"_{"n"}is the sum of the numbers divided by "n"::$frac\{x\_1\; +\; x\_2\; +\; cdots\; +\; x\_n\}\{n\}.$

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:

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

If "x"

_{1}, "x"_{2}, . . ., "x"_{"n"}> 0, this is equal to the exponential of the arithmetic mean of thenatural logarithm s of the numbers::$exp\; left(\; frac\{ln\; \{x\_1\}\; +\; ln\; \{x\_2\}\; +\; cdots\; +\; ln\; \{x\_n\{n\}\; ight).$

**The inequality**Restating the inequality using mathematical notation, we have that for any list of "n" nonnegative real numbers "x"

_{1}, "x"_{2}, . . ., "x"_{"n"},:$frac\{x\_1\; +\; x\_2\; +\; cdots\; +\; x\_n\}\{n\}\; geq\; sqrt\; [n]\; \{x\_1\; cdot\; x\_2\; cdots\; x\_n\},$

and that if and only if "x"

_{1}= "x"_{2}= . . . = "x"_{"n"},:$frac\{x\_1\; +\; x\_2\; +\; cdots\; +\; x\_n\}\{n\}\; =\; sqrt\; [n]\; \{x\_1\; cdot\; x\_2\; cdots\; x\_n\}.$

**Generalizations**There is a similar inequality for the

weighted arithmetic mean andweighted geometric mean . Specifically, let the nonnegative numbers "x"_{1}, "x"_{2}, . . ., "x"_{"n"}and the nonnegative weights "α"_{1}, "α"_{2}, . . ., "α"_{"n"}be given. Set $alpha\; =\; alpha\_1\; +\; alpha\_2\; +\; cdots\; +\; alpha\_n$. If "α" > 0, then the inequality:$frac\{alpha\_1\; x\_1\; +\; alpha\_2\; x\_2\; +\; cdots\; +\; alpha\_n\; x\_n\}\{alpha\}\; geq\; sqrt\; [alpha]\; \{x\_1^\{alpha\_1\}\; x\_2^\{alpha\_2\}\; cdots\; x\_n^\{alpha\_n$

holds with equality if and only if all the "x

_{k}" with "α_{k}" > 0 are equal. Here the convention 0^{0}= 1 is used.If all "α

_{k}" = 1, this reduces to the above AM-GM inequality.Other generalizations of the inequality of arithmetic and geometric means are given by

Muirhead's inequality ,MacLaurin's inequality , and the generalized mean inequality.**Example application**Consider the following function:

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

for "x", "y", and "z" all positive real numbers. Suppose we wish to find the minimum value of this function. Rewriting a bit, and applying the AM-GM inequality, we have:

:

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

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

**Proof by induction**There are several ways to prove the AM-GM inequality; for example, it can be inferred from

Jensen's inequality , using the concave function ln("x"). It can also be proven using therearrangement inequality . Considering length and required prerequisites, the proof by induction given below is probably the best recommendation for first reading.With the arithmetic mean:$mu=frac\{\; x\_1\; +\; cdots\; +\; x\_n\}n$of the non-negative real numbers "x"

_{1},...,"x_{n}", the AM-GM statement is equivalent to:$mu^nge\; x\_1\; x\_2\; cdots\; x\_n,$with equality if and only if "μ" = "x_{i}" for all "i" = 1,...,"n".For the following proof we apply

mathematical induction and only well-known rules of arithmetic.**Induction basis:**For "n" = 1 the statement is true with equality.**Induction hypothesis:**Suppose that the AM-GM statement holds for all choices of "n" non-negative real numbers.**Induction step:**Consider "n" + 1 non-negative real numbers. Their arithmetic mean "μ" satisfies:$(n+1)mu=\; x\_1\; +\; cdots\; +\; x\_n\; +\; x\_\{n+1\}.,$If all numbers are equal to "μ", then we have equality in the AM-GM statement and we are done. Otherwise we may find one number that is greater than "μ" and one that is smaller than "μ", say "x_{n}" > "μ" and "x"_{"n"+1}< "μ". Then:$(x\_n-mu)(mu-x\_\{n+1\})>0,.qquad(*)$

Now consider the "n" numbers:$x\_1,\; ldots,\; x\_\{n-1\},\; x\_n\text{'}$ with $x\_n\text{'}:=x\_n+x\_\{n+1\}-muge\; x\_n-mu>0,,$

which are also non-negative. Since

:$nmu=x\_1\; +\; cdots\; +\; x\_\{n-1\}\; +\; underbrace\{x\_n+x\_\{n+1\}-mu\}\_\{=,x\_n\text{'}\},$

"μ" is also the arithmetic mean of $x\_1,\; ldots,\; x\_\{n-1\},\; x\_n\text{'}$ and the induction hypothesis implies

:$mu^\{n+1\}=mu^ncdotmuge\; x\_1x\_2\; cdots\; x\_\{n-1\}\; x\_n\text{'}mu.qquad(**)$

Due to (*) we know that

:$(underbrace\{x\_n+x\_\{n+1\}-mu\}\_\{=,x\_n\text{'}\})mu-x\_nx\_\{n+1\}=(x\_n-mu)(mu-x\_\{n+1\})>0,$

hence

:$x\_n\text{'}mu>x\_nx\_\{n+1\},,qquad(\{*\}\{*\}\{*\})$

in particular "μ" > 0. Therefore, if at least one of the numbers "x"

_{1},...,"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 (***). Therefore, substituting (***) into (**) gives in both cases:$mu^\{n+1\}>x\_1x\_2\; cdots\; x\_\{n-1\}\; x\_nx\_\{n+1\},,$

which completes the proof.

**Proof by Pólya**George Pólya provided the following proof, using theexponential function and the inequality "e"^{"x"}≥ 1 + "x", which is valid for every real number "x". To verify this inequality, note that both sides as well as their firstderivative s agree for "x" = 0 and that the exponential function is strictly convex, because its second derivative is positive for every real "x". For this reason, also the equality "e"^{"x"}= 1 + "x" only holds for "x" = 0.Let μ be the arithmetic mean, and let ρ be the geometric mean of "x"

_{1}, ..., "x"_{"n"}. If all "x"_{1}, ..., "x"_{"n"}are equal, then μ = ρ.It remains to prove the strict inequality μ > ρ if "x"

_{1}, ..., "x"_{"n"}≥ 0 are not all equal. Then, in particular, they are not all zero, hence μ > 0.If we substitute "x"

_{"i"}/μ − 1 for "x" in the above inequality "e"^{"x"}≥ 1 + "x" we get that:$expBigl(\{x\_i\; over\; mu\}\; -\; 1Bigr)ge\; \{x\_i\; over\; mu\},$

for each "i" and strict inequality for those "i" with "x"

_{"i"}≠ μ. Since "x"_{"i"}/μ ≥ 0, we can multiply all these inequalities together, side by side, for "i" = 1, ..., "n", to obtain:$prod\_\{i=1\}^nexpBigl(\{x\_i\; over\; mu\}\; -\; 1Bigr)\; >\; prod\_\{i=1\}^n\; \{x\_i\; over\; mu\},,$

where we get strict inequality because no factor on the left-hand side is zero and there was strict inequality for at least one "i". Using the

functional equation of the exponential function, we get:$expiggl(frac1musum\_\{i=1\}^n\; x\_i\; -\; niggr)\; prod\_\{i=1\}^n\; \{x\_i\; over\; mu\},.qquad(*)$

Since μ is the arithmetic mean, the summation in the parentheses on the left of (*) can be reduced to

:$sum\_\{i=1\}^n\; x\_i\; =\; nmu,.$

Thus, the left-hand side of the inequality (*) is exp("n" − "n") = 1. Since ρ is the geometric mean, the product on the right of (*) can be rewritten as

:$frac1\{mu^n\}prod\_\{i=1\}^n\; x\_i\; =\; \{\; ho^n\; over\; mu^n\}.$

So (*) reduces to 1 > ρ

^{"n"}/μ^{"n"}and hence μ > ρ.**Proof by Cauchy**The following proof by cases relies directly on well-known rules of arithmetic but employs the rarely-used technique of forward-backward-induction. It is essentially from

Augustin Louis Cauchy and can be found in his "Cours d'analyse".**The case where all the terms are equal**If all the terms are equal:

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

then their sum is "nx"

_{1}, so their arithmetic mean is "x"_{1}; and their product is "x"_{1}^{"n"}, so their geometric mean is "x"_{1}; therefore, the arithmetic mean and geometric mean are equal, as desired.**The case where not all the terms are equal**It remains to show that if "not" all the terms are equal, then the arithmetic mean is greater than the geometric mean. Clearly, this is only possible when "n" > 1.

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

= The subcase where "n"= 2 =If "n" = 2, then we have two terms, "x"

_{1}and "x"_{2}, and since (by our assumption) not all terms are equal, we have:: $egin\{align\}x\_1\; e\; x\_2\; \backslash \; [3pt]\; x\_1\; -\; x\_2\; e\; 0\; \backslash \; [3pt]\; left(\; x\_1\; -\; x\_2\; ight)\; ^2\; 0\; \backslash \; [3pt]\; x\_1^2\; -\; 2\; x\_1\; x\_2\; +\; x\_2^2\; 0\; \backslash \; [3pt]\; x\_1^2\; +\; 2\; x\_1\; x\_2\; +\; x\_2^2\; 4\; x\_1\; x\_2\; \backslash \; [3pt]\; left(\; x\_1\; +\; x\_2\; ight)\; ^2\; 4\; x\_1\; x\_2\; \backslash \; [3pt]\; Bigl(\; frac\{x\_1\; +\; x\_2\}\{2\}\; Bigr)^2\; x\_1\; x\_2\; \backslash \; [3pt]\; frac\{x\_1\; +\; x\_2\}\{2\}\; sqrt\{x\_1\; x\_2\}end\{align\}$

as desired.

= The subcase where "n"= 2^{"k"}=Consider the case where "n" = 2

^{"k"}, where "k" is a positive integer. We proceed bymathematical induction .In the base case, "k" = 1, so "n" = 2. We have already shown that the inequality holds where "n" = 2, so we are done.

Now, suppose that for a given "k" > 1, we have already shown that the inequality holds for "n" = 2

^{"k"−1}, and we wish to show that it holds for "n" = 2^{"k"}. To do so, we proceed as follows:: $egin\{align\}frac\{x\_1\; +\; x\_2\; +\; cdots\; +\; x\_\{2^k\{2^k\}\; \{\}\; =frac\{frac\{x\_1\; +\; x\_2\; +\; cdots\; +\; x\_\{2^\{k-1\}\{2^\{k-1\; +\; frac\{x\_\{2^\{k-1\}\; +\; 1\}\; +\; x\_\{2^\{k-1\}\; +\; 2\}\; +\; cdots\; +\; x\_\{2^k\{2^\{k-1\}\{2\}\; \backslash \; [7pt]\; ge\; frac\{sqrt\; [2^\{k-1\}]\; \{x\_1\; cdot\; x\_2\; cdots\; x\_\{2^\{k-1\}\; +\; sqrt\; [2^\{k-1\}]\; \{x\_\{2^\{k-1\}\; +\; 1\}\; cdot\; x\_\{2^\{k-1\}\; +\; 2\}\; cdots\; x\_\{2^k\}\{2\}\; \backslash \; [7pt]\; ge\; sqrt\{sqrt\; [2^\{k-1\}]\; \{x\_1\; cdot\; x\_2\; cdots\; x\_\{2^\{k-1\}\; cdot\; sqrt\; [2^\{k-1\}]\; \{x\_\{2^\{k-1\}\; +\; 1\}\; cdot\; x\_\{2^\{k-1\}\; +\; 2\}\; cdots\; x\_\{2^k\}\; \backslash \; [7pt]\; =\; sqrt\; [2^k]\; \{x\_1\; cdot\; x\_2\; cdots\; x\_\{2^kend\{align\}$

where in the first inequality, the two sides are equal only if both of the following are true:

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

(in which case the first arithmetic mean and first geometric mean are both equal to "x"

_{1}, and similarly with the second arithmetic mean and second geometric mean); and in the second inequality, the two sides are only equal if the two geometric means are equal. Since not all 2^{"k"}numbers are equal, it is not possible for both inequalities to be equalities, so we know that::$frac\{x\_1\; +\; x\_2\; +\; cdots\; +\; x\_\{2^k\{2^k\}\; >\; sqrt\; [2^k]\; \{x\_1\; cdot\; x\_2\; cdots\; x\_\{2^k$

as desired.

**The subcase where "n" < 2**^{"k"}If "n" is not a natural power of 2, then it is certainly "less" than some natural power of 2, since the sequence 2, 4, 8, . . ., 2

^{"k"}, . . . is unbounded above. Therefore, without loss of generality, let "m" be some natural power of 2 that is greater than "n".So, if we have "n" terms, then let us denote their arithmetic mean by α, and expand our list of terms thus:

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

We then have:

: $egin\{align\}alpha\; =\; frac\{x\_1\; +\; x\_2\; +\; cdots\; +\; x\_n\}\{n\}\; \backslash \; [6pt]\; =\; frac\{frac\{m\}\{n\}\; left(\; x\_1\; +\; x\_2\; +\; cdots\; +\; x\_n\; ight)\}\{m\}\; \backslash \; [6pt]\; =\; frac\{x\_1\; +\; x\_2\; +\; cdots\; +\; x\_n\; +\; frac\{m-n\}\{n\}\; left(\; x\_1\; +\; x\_2\; +\; cdots\; +\; x\_n\; ight)\}\{m\}\; \backslash \; [6pt]\; =\; frac\{x\_1\; +\; x\_2\; +\; cdots\; +\; x\_n\; +\; left(\; m-n\; ight)\; alpha\}\{m\}\; \backslash \; [6pt]\; =\; frac\{x\_1\; +\; x\_2\; +\; cdots\; +\; x\_n\; +\; x\_\{n+1\}\; +\; cdots\; +\; x\_m\}\{m\}\; \backslash \; [6pt]\; sqrt\; [m]\; \{x\_1\; cdot\; x\_2\; cdots\; x\_n\; cdot\; x\_\{n+1\}\; cdots\; x\_m\}\; \backslash \; [6pt]\; =\; sqrt\; [m]\; \{x\_1\; cdot\; x\_2\; cdots\; x\_n\; cdot\; alpha^\{m-n,,end\{align\}$

so

: $egin\{align\}alpha^m\; x\_1\; cdot\; x\_2\; cdots\; x\_n\; cdot\; alpha^\{m-n\}\; \backslash \; [3pt]\; alpha^n\; x\_1\; cdot\; x\_2\; cdots\; x\_n\; \backslash \; [3pt]\; alpha\; sqrt\; [n]\; \{x\_1\; cdot\; x\_2\; cdots\; x\_n\}end\{align\}$

as desired.

**Proof of the generalized AM-GM inequality using Jensen's inequality**Using the finite form of

Jensen's inequality for thenatural logarithm , we can prove the inequality between the weighted arithmetic mean and the weighted geometric mean stated above.Since an "x

_{k}" with weight "α_{k}" = 0 has no influence on the inequality, we may assume in the following that all weights are positive. If all "x_{k}" are equal, then equality holds. Therefore, it remains to prove strict inequality if they are not all equal, which we will assume in the following, too. If at least one "x_{k}" is zero (but not all), then the weighted geometric mean is zero, while the weighted arithmetic mean is positive, hence strict inequality holds. Therefore, we may assume also that all "x_{k}" are positive.Since the natural logarithm is strictly concave, the finite form of Jensen's inequality and the

functional equation s of the natural logarithm imply:$lniggl(frac\{alpha\_1x\_1+cdots+alpha\_nx\_n\}alphaiggr)frac\{alpha\_1\}alphaln\; x\_1+cdots+frac\{alpha\_n\}alphaln\; x\_n=ln\; sqrt\; [alpha]\; \{x\_1^\{alpha\_1\}\; x\_2^\{alpha\_2\}\; cdots\; x\_n^\{alpha\_n.$Since the natural logarithm is strictly increasing,:$frac\{alpha\_1x\_1+cdots+alpha\_nx\_n\}alpha>sqrt\; [alpha]\; \{x\_1^\{alpha\_1\}\; x\_2^\{alpha\_2\}\; cdots\; x\_n^\{alpha\_n.$

**References*** Augustin-Louis Cauchy, [

*http://visualiseur.bnf.fr/Visualiseur?Destination=Gallica&O=NUMM-29058 "Cours d'analyse de l'École Royale Polytechnique, premier partie, Analyse algébrique,"*] Paris, 1821. The proof of the inequality of arithmetic and geometric means can be found on pages 457ff.

*Wikimedia Foundation.
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