Inequality of Arithmetic and Geometric Means

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## Background

## The inequality

## Geometric interpretation

## Example application

## Practical applications

## Proofs of the AM-GM inequality

### Proof using Jensen's inequality

### Proofs by induction

#### Proof by induction #1

#### Proof by induction #2

### Proof by Cauchy using forward-backward induction

#### The case where all the terms are equal

#### The case where not all the terms are equal

##### The subcase where *n* = 2

##### The subcase where *n* = 2^{k}

##### The subcase where *n* < 2^{k}

### Proof by induction using basic calculus

### Proof by Pólya using the exponential function

### Proof by Lagrangian Multipliers

## Generalizations

### Weighted AM-GM inequality

### Proof using Jensen's inequality

### Matrix Arithmetic Geometric Mean Inequality

### Other generalizations

## See also

## Notes

## References

## External links

This article uses material from the Wikipedia page available here. It is released under the Creative Commons Attribution-Share-Alike License 3.0.

Inequality of Arithmetic and Geometric Means

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 numbers 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.

The simplest non-trivial case -- i.e., with more than one variable -- for two non-negative numbers x and y, is the statement that

with equality if and only if *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 (*a* ± *b*)^{2} = *a*^{2} ± 2*ab* + *b*^{2} of the binomial formula:

Hence (*x* + *y*)^{2} >= 4*xy*, 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 by *2*.

For a geometrical interpretation, consider a rectangle with sides of length x and y, hence it has perimeter 2*x* + 2*y* and area xy. Similarly, a square with all sides of length has the perimeter 4 and the same area as the rectangle. The simplest non-trivial case of the AM-GM inequality implies for the perimeters that 2*x* + 2*y* >= 4 and that only the square has the smallest perimeter amongst all rectangles of equal area.

Extensions of the AM-GM inequality are available to include weights or generalized means.

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:

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:

If *x*_{1}, *x*_{2}, . . . , *x _{n}* > 0, this is equal to the exponential of the arithmetic mean of the natural logarithms of the numbers:

Restating the inequality using mathematical notation, we have that for any list of n nonnegative real numbers *x*_{1}, *x*_{2}, . . . , *x _{n}*,

and that equality holds if and only if *x*_{1} = *x*_{2} = · · · = *x _{n}*.

In two dimensions, 2*x*_{1} + 2*x*_{2} is the perimeter of a rectangle with sides of length *x*_{1} and *x*_{2}. Similarly, 4 is the perimeter of a square with the same area, *x*_{1}*x*_{2}, as that rectangle. Thus for *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 n dimensions. Every vertex of an n-dimensional box is connected to n edges. If these edges' lengths are *x*_{1}, *x*_{2}, . . . , *x _{n}*, then

is the total length of edges connected to a vertex on an n-dimensional cube of equal volume, since in this case *x*_{1}=...=*x*_{n}. Since the inequality says

it can be restated by multiplying through by *n*2^{n-1} to obtain

with equality if and only if
*x*_{1} = *x*_{2} = · · · = *x _{n}*.

Thus the AM-GM inequality states that only the n-cube has the smallest sum of lengths of edges connected to each vertex amongst all n-dimensional boxes with the same volume.^{[2]}

Consider the function

for all positive real numbers x, y and z. Suppose we wish to find the minimal value of this function. First we rewrite it a bit:

with

Applying the AM-GM inequality for *n* = 6, we get

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

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

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 investments, as the average return overstates the cumulative effect.

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

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

We have to show that

with equality only when all numbers are equal. If *x _{i}* ?

Thus the right-hand side will be largest when all x_{i}s are equal to the arithmetic mean

thus as this is then the largest value of right-hand side of the expression, we have

This is a valid proof for the case *n* = 2, but the procedure of taking iteratively pairwise averages may fail to produce n equal numbers in the case *n* >= 3. An example of this case is *x*_{1} = *x*_{2} ? *x*_{3}: Averaging two different numbers produces two equal numbers, but the third one is still different. Therefore, we never actually get an inequality involving the geometric mean of three equal numbers.

Hence, an additional trick or a modified argument is necessary to turn the above idea into a valid proof for the case *n* >= 3.

Of the non-negative real numbers *x*_{1}, . . . , *x _{n}*, the AM-GM statement is equivalent to

with equality if and only if *?* = *x _{i}* for all

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 *x*_{1}, . . . , *x*_{n+1}, . Their arithmetic mean ? satisfies

If all the x_{i} are equal to ?, then we have equality in the AM-GM statement and we are done. In the case where some are not equal to ?, there must exist one number that is greater than the arithmetic mean ?, and one that is smaller than ?. Without loss of generality, we can reorder our x_{i} in order to place these two particular elements at the end: *x _{n}* >

Now define *y* with

and consider the n numbers *x*_{1}, . . . , *x*_{n-1}, y which are all non-negative. Since

Thus, ? is also the arithmetic mean of n numbers *x*_{1}, . . . , *x*_{n-1}, *y* and the induction hypothesis implies

Due to (*) we know that

hence

in particular *?* > 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 (***) to get a lower bound of the right-hand side of (**). Thus, in both cases we can substitute (***) into (**) to get

which completes the proof.

First of all we shall prove that for real numbers *x*_{1} < 1 and *x*_{2} > 1 there follows

Indeed, multiplying both sides of the inequality *x*_{2} > 1 by 1 - *x*_{1}, gives

whence the required inequality is obtained immediately.

Now, we are going to prove that for positive real numbers *x*_{1}, . . . , *x*_{n} satisfying
*x*_{1} . . . *x*_{n} = 1, there holds

The equality holds only if *x*_{1} = ... = *x*_{n} = 1.

**Induction basis:** For *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 *n* - 1.

**Induction step:** Consider natural number *n*, i.e. for positive real numbers *x*_{1}, . . . , *x*_{n}, there holds *x*_{1} . . . *x*_{n} = 1. There exists at least one *x _{k}* < 1, so there must be at least one

Further, the equality *x*_{1} . . . *x*_{n} = 1 we shall write in the form of (*x*_{1} . . . *x*_{n-2}) (*x*_{n-1}*x*_{n}) = 1. Then, the induction hypothesis implies

However, taking into account the induction basis, we have

which completes the proof.

For positive real numbers *a*_{1}, . . . , *a*_{n}, let's denote

The numbers *x*_{1}, . . . , *x*_{n} satisfy the condition *x*_{1} . . . *x*_{n} = 1. So we have

whence we obtain

with the equality holding only for *a*_{1} = ... = *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*.^{[3]}

If all the terms are equal:

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.

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.

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:

hence

as desired.

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

In the base case, *k* = 1, so *n* = 2. We have already shown that the inequality holds when *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 apply the inequality twice for 2^{k-1} numbers and once for 2 numbers to obtain:

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

and

(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:

as desired.

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 ?, and expand our list of terms thus:

We then have:

so

and

as desired.

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

**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**: In order to prove the statement for *n* + 1 non-negative real numbers *x*_{1}, . . . , *x _{n}*,

with equality only if all the *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 *n* + 1 numbers are positive.

We consider the last number *x*_{n+1} as a variable and define the function

Proving the induction step is equivalent to showing that *f*(*t*) >= 0 for all *t* > 0, with *f*(*t*) = 0 only if *x*_{1}, . . . , *x _{n}* and t are all equal. This can be done by analyzing the critical points of f using some basic calculus.

The first derivative of f is given by

A critical point *t*_{0} has to satisfy *f?*(*t*_{0}) = 0, which means

After a small rearrangement we get

and finally

which is the geometric mean of *x*_{1}, . . . , *x _{n}*. This is the only critical point of f. Since

where the final inequality holds due to the induction hypothesis. The hypothesis also says that we can have equality only when *x*_{1}, . . . , *x _{n}* are all equal. In this case, their geometric mean

This technique can be used in the same manner to prove the generalized AM-GM inequality and Cauchy-Schwarz inequality in Euclidean space **R**^{n}.

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

Consider a list of non-negative real numbers *x*_{1}, *x*_{2}, . . . , *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

with equality if and only if *x _{i}* =

Returning to (*),

which produces *x*_{1}*x*_{2} · · · *x _{n}* ?

If any of the are , then there is nothing to prove. So we may assume all the are strictly positive.

Because the arithmetic and geometric means are homogeneous of degree 1, without loss of generality assume that . Set , and . The inequality will be proved (together with the equality case) if we can show that the minimum of subject to the constraint is equal to , and the minimum is only achieved when . Let us first show that the constrained minimization problem has a global minimum.

Set . Since the intersection is compact, the extreme value theorem guarantees that the minimum of subject to the constraints and is attained at some point inside . On the other hand, observe that if any of the , then , while , and . This means that the minimum inside is in fact a global minimum, since the value of at any point inside is certainly no smaller than the minimum, and the value of at any point not inside is strictly bigger than the value at , which is no smaller than the minimum.

The method of Lagrange multipliers says that the global minimum is attained at a point where the gradient of is times the gradient of , for some . We will show that the only point at which this happens is when and

Compute and

along the constraint. Setting the gradients proportional to one another therefore gives for each that and so Since the left-hand side does not depend on , it follows that , and since , it follows that and , as desired.

There is a similar inequality for the weighted arithmetic mean and weighted geometric mean. Specifically, let the nonnegative numbers *x*_{1}, *x*_{2}, . . . , *x _{n}* and the nonnegative weights

holds with equality if and only if all the x_{k} with *w _{k}* > 0 are equal. Here the convention 0

If all *w _{k}* = 1, this reduces to the above inequality of arithmetic and geometric means.

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 x_{k} with weight *w _{k}* = 0 has no influence on the inequality, we may assume in the following that all weights are positive. If all x

Since the natural logarithm is strictly concave, the finite form of Jensen's inequality and the functional equations of the natural logarithm imply

Since the natural logarithm is strictly increasing,

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 and are positive semi-definite the matrix may not be positive semi-definite and hence may not have a canonical square root. In ^{[5]} Bhatia and Kittaneh proved that for any unitarily invariant norm and positive semi-definite matrices and it is the case that

Later, in ^{[6]} the same authors proved the stronger inequality that

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

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

New inequalities with classical means appeared in a series of publications (see [7]).

**^**Hoffman, D. G. (1981), "Packing problems and inequalities", in Klarner, David A. (ed.),*The Mathematical Gardner*, Springer, pp. 212-225, doi:10.1007/978-1-4684-6686-7_19**^**Steele, J. Michael (2004).*The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities*. MAA Problem Books Series. Cambridge University Press. ISBN 978-0-521-54677-5. OCLC 54079548.**^**Cauchy, Augustin-Louis (1821).*Cours d'analyse de l'École Royale Polytechnique, première partie, Analyse algébrique,*Paris. The proof of the inequality of arithmetic and geometric means can be found on pages 457ff.**^**Arnold, Denise; Arnold, Graham (1993).*Four unit mathematics*. Hodder Arnold H&S. p. 242. ISBN 978-0-340-54335-1. OCLC 38328013.**^**divy patel, Rajendra; Kittaneh, Fuad (1990). "On the singular values of products of operators".*SIAM Journal of Matrix Analysis*.**11**(2): 272-277. doi:10.1137/0611018.**^**Bhatia, Rajendra; Kittaneh, Fuad (2000). "Notes on matrix arithmetic-geometric mean inequalities".*Linear Algebra and Its Applications*.**308**(1-3): 203-211. doi:10.1016/S0024-3795(00)00048-3. Retrieved 2020.**^**If AC =*a*and BC =*b*. OC =**AM**of*a*and*b*, and radius*r*= QO = OG.

Using Pythagoras' theorem, QC² = QO² + OC² ∴ QC = √QO² + OC² =**QM**.

Using Pythagoras' theorem, OC² = OG² + GC² ∴ GC = √OC² − OG² =**GM**.

Using similar triangles, = ∴ HC = =**HM**.

7. Florin Nichita, On Classical Means Inequalities, Scholarly Community Encyclopedia, MDPI, https://encyclopedia.pub/2364 - Created: 20 Aug 2020; Latest updated: 20 Aug 2020

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

This article uses material from the Wikipedia page available here. It is released under the Creative Commons Attribution-Share-Alike License 3.0.

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