The geometric mean of a non-empty data set of (positive) numbers **is always at most their arithmetic mean**. Equality is only obtained when all numbers in the data set are equal; otherwise, the geometric mean is smaller. For example, the geometric mean of 2 and 3 is 2.45, while their arithmetic mean is 2.5.

Contents

- 1 What is relation between arithmetic mean and geometric mean?
- 2 When am GM and HM are equal?
- 3 Is the arithmetic mean always greater than the geometric mean?
- 4 What is the relation between arithmetic mean am and geometric mean GM?
- 5 What is meant by arithmetic mean geometric mean and harmonic mean state the relation between these three means justify the claim?
- 6 What is arithmetic mean and harmonic mean?
- 7 What is the difference between arithmetic and geometric?
- 8 Which is better arithmetic or geometric mean?
- 9 What is the difference between arithmetic and geometric returns?
- 10 Arithmetic Mean – Geometric Mean
- 11 Art of Problem Solving
- 12 Proofs
- 13 Generalizations
- 14 Problems
- 15 Geometric Mean vs Arithmetic Mean
- 16 The Difference Between the Arithmetic Mean and Geometric Mean
- 17 The Formula for Arithmetic Average
- 18 How to Calculate the Arithmetic Average
- 19 The Formula for Geometric Average
- 20 How to Calculate the Geometric Average
- 21 Using the Arithmetic Mean-Geometric Mean Inequality in Problem Solving.
- 22 Some algebra
- 23 Another Geometry Example
- 24 Cost of Fencing a Field
- 25 Minimum of
- 26 Construct a Square with Same Area as a Given Rectangle
- 27 Maximum Area of a Sector of a Circle With Fixed Perimeter
- 28 Sector as Fraction of a Circle
- 29 Sector in Radian Measure
- 30 Inequalities Problems
- 31 Maximumand Minimum of
- 32 Maximum off(x) = (1-x)(1+x)(1+x)
- 33 MinimumSurface Area of a Can of Fixed Volume
- 34 Maximum area of a Pen, fixed length of fencing, with Partitions Parallel to a Side
- 35 The Arithmetic-Geometric Mean Inequality
- 36 Proof of the inequality
- 37 Demonstration: Turning hard calculus problems into easy inequality problems
- 38 Conclusion
- 39 Copyright stuff

## What is relation between arithmetic mean and geometric mean?

Let A and G be the Arithmetic Means and Geometric Means respectively of two positive numbers m and n. Then, we have A = m + n/2 and G = ±√mn. Since, m and n are positive numbers, hence it is evident that A > G when G = -√mn.

## When am GM and HM are equal?

The product of arithmetic mean and harmonic mean is equal to the square of the geometric mean. AM × HM = GM^{2}. Among the three means, the arithmetic mean is greater than the geometric mean, and the geometric mean is greater than the harmonic mean.

## Is the arithmetic mean always greater than the geometric mean?

The arithmetic mean is always higher than the geometric mean as it is calculated as a simple average. It is applicable only to only a positive set of numbers. It can be calculated with both positive and negative sets of numbers. Geometric mean can be more useful when the dataset is logarithmic.

## What is the relation between arithmetic mean am and geometric mean GM?

If AM and GM are the arithmetic mean and the geometric mean of two positive integers a and b, respectively, then, AM>GM. Hence proved that the arithmetic mean of two positive numbers is always greater than their GM. This is also called the arithmetic mean – geometric mean (AM-GM) inequality.

## What is meant by arithmetic mean geometric mean and harmonic mean state the relation between these three means justify the claim?

Harmonic mean has the least value among all the three means. The relationship between arithmetic mean, geometric mean and harmonic mean is: “ The product of arithmetic mean and harmonic mean of any two numbers a and b in such a way that a > b > 0 is equal to the square of their geometric mean.”

## What is arithmetic mean and harmonic mean?

An arithmetic average is the sum of a series of numbers divided by the count of that series of numbers. The harmonic mean is best used for fractions such as rates or multiples.

## What is the difference between arithmetic and geometric?

An arithmetic sequence has a constant difference between each consecutive pair of terms. A geometric sequence has a constant ratio between each pair of consecutive terms. This would create the effect of a constant multiplier.

## Which is better arithmetic or geometric mean?

The arithmetic mean is more useful and accurate when it is used to calculate the average of a data set where numbers are not skewed and not dependent on each other. However, in the scenario where there is a lot of volatility in a data set, a geometric mean is more effective and more accurate.

## What is the difference between arithmetic and geometric returns?

Arithmetic returns are the everyday calculation of the average. The geometric mean is calculated by multiplying all the (1+ returns), taking the n-th root and subtracting the initial capital (1). The result is the same as compounding the returns across the years.

## Arithmetic Mean – Geometric Mean

The Arithmetic Mean – Geometric Mean inequality, sometimes known as the AM-GM inequality, is defined as follows: The geometric mean cannot be greater than the arithmetic mean, and they will be equal if and only if all of the numbers picked are equal in the first instance. That is, a1+a2+anna1a2.annfracgesqrtwith equality if and only ifa1=a2=ana 1=a 2=cdots =a n, and a1=a2+anna1a2.annfracgesqrtwith equality if and only ifa1=a2=ana 1=a 2=cdots =a n To put it another way, i=1nain i=1nain. ge _sqrt Take, for example, the scenario where n=2,n=2, i.e.

The AM-GM inequality therefore saysx+y2xy.dfracge xy.dfracge xy.dfracge sqrt.

It is also possible to have equality whenx=ysqrt x=sqrt yor if the variables are equal to each other.

Inducting variables based on the number of variables is a frequent strategy.

Given the AM-GM disparity, here is a straightforward illustration.

Allow the two numbers to beaa and bb to be used.

The AM-GM inequality asserts thata+b2ab,dfracgeq sqrt,implying for this issue thata+b2100=20.a + bgeq 2 sqrt= 20.a + bgeq 2 sqrt= 20.a + bgeq 2 sqrt= 20.a + bgeq 2 sqrt= 20.a + bgeq 2 sqrt= 20.a + bgeq 2 sqrt Note:When a=b=10a = b=10a = b=10a = b=10a = b=10a = b=10a = b=10a = b=10a = b=10a = b=10a = b=10a = b=10a = b=10a = b=10a = b=10 You can experiment with an issue similar to the one shown above: The sum of two positive real numbers equals one hundred.

- Find the most profitable product for them.
- The three geen jellies are a trio of geen jellies.
- There is no way to establish a definitive relationship.
- The three red cubes have side lengths of abc, a b, and c, while the three green cuboids have dimensions of abc, a times b times c, as illustrated in the diagram above.
- AM-GM inequality may be demonstrated using a variety of approaches.
- The first item on the list is to demonstrate using some type of inductive reasoning.

Conventional inductions examine a base case and then demonstrate thatP(n)P(n+1)P(n)P(n)P(n)P(n+1)P(n)P(n)P(n)P(n)P(n)P(n)P(n)P(n)P(n)P(n)P(n)P(n)P(n)P(n)P(n) However, we will demonstrate the following:

- P(2)P(2)holds
- sP(n)⟹P(2n). P(n) implies P(2n)
- P(n) implies P(n1). P(n) implies P(n1). P(n) is equivalent to P(n-1)

What makes you think this will work? It is important to note that we leap fromnnto2n2nby demonstrating thatP(n)P(2n)P(n)implies P(n) (2n). In this case, we may induct backwards from2n2nton+1,n+1, usingP(n)P(n1)P(n)implies P(n-1), in order to check that all integers in the range of nand 2n2n(inclusive) fulfill the claim. Forward-backward induction is the term used to describe this. We’ll now proceed to demonstrate our points. It has previously been demonstrated that P(2)P(2) holds, and now we will demonstrate that P(n)P(2n)P(n)implies P.

- Positive realsa1, A2,.,a2na 1, A 2,.,a_ are taken into consideration.
- For anynnpositive reals, we can assume thatP(n)P(n)is true.
- In addition, we’ve utilized the base example, which has 22 variables.
- It is still necessary to demonstrate that P(n)P(n1)P(n)implies P(n) (n-1).
- Then you’ll observe that 1ni=1nai=1n(i=1n1ai+1n1i=1n1ai)=1n1i=1n1ai=1n(i=1n1ai+1n1i=1n1ai)=1n1i=1n1ai.
- dfracsum_ n a i =dfracleft(sum_ a i + dfracsum_ a iright)=dfracsum_ a i.
- As a result, the third element of the evidence has been completed, as has the induction.
- So let’s use that method to solving the problem.
- f(x) = log x for every x0 is defined as f(x).
- Thus, f(x)f(x) turns out to be a concave function when x = 1.

As a result, according to Jensen’s inequality, we havef(i=1nain)i=1nf(ai)nlog(i=1nain)i=1nlog(ai)n.beginfleft(dfrac n a i)right)f(i=1nain)f(i=1nain)f(i=1nain)f(i=1nain)f(i= eq dfrac n f left(a I right) eq dfrac n f left(a I right) eq dfrac n f left(a I right) eq dfrac n f left(a I right) eq dfrac n f left(a I right) eq dfrac n f left(a I right Rightarrow dfrac n a i dfrac n a i dfrac n a i dfrac n a i dfrac n a i dfrac n a i dfrac n a i dfrac n a i dfrac n a i dfrac n a i eq dfrac n log a I right eq dfrac n log a I right eq dfrac n log a I right eq dfrac n log a I right eq dfrac n log a I right eq dfrac n log a I right eq dfrac n log a I right eq dfrac n In accordance with the property of logarithms, we have the following: logax+logay=logabloga = log ab and bloga=logab log x + log y = log log xy.

In order to simplify the terms on the RHS, we can write them as log(i=1nain)i=1nlogain=log(i=1nain)i=1/n.beginlog left(dfrac n a i)geq dfrac n a i = dfrac n = log(i=1nain)1/n.endlog (i=1nain)i=1nlogain=log Another demonstration, made famous by a mathematician named George Pólya, does not rely on inductive reasoning to establish its validity.

Assume that given a sequence of positive realsaka kwith1kn1leq k nand a sequence of positive realspkp kwith1kn1leq k nsuch thatk=1npk=1sum_ p k = 1, a1p1a2p2a3p3anpna1p1a2p2a3p3+anpn.a 1p1a2p2a3p First, Pólya makes the observation that 1+xex,1+xleq ex, which can be readily proven visually by noting that equality occurs only atx=0x=0.

- This idea, according to legend, came to Pólya in a dream, in which he exclaimed that it was “the greatest mathematics he had ever imagined.” If we make the adjustment in variablesxx1xmapsto x-1, our initial observation becomesxex1xleq e, and our final observation becomesxex1xleq e.
- As a result, we may sayA,Gexp(k=1nakpk1).
- We have connected AA and GG via inequality, but we have not distinguished between them.
- The concept of “normalization” comes to mind at this point.

Create a new sequence k alpha k with 1 k n, 1 leq k n, where we have k=akAA=a1p1+a2p2+a3p3+anpn.begin alpha k= frac A =a 1p 1+a2p2+a3p3+cdots+a np n.end alpha k= frac A =a 1p 1+ Using our previously established bound forGGfor our new variableskalpha k, we obtain k=1nkpkexp(k=1nkpk1)k=1n(akA)pkexp(k=1nakApk1)k=1n(akA)pkexp(k=1nakApk1)k=1n(akA)pkexp(k=1nakApk1)k=1n(akA) beginprod_ alpha kleq exp left(sum_ alpha kp k-1 right) beginprod_ alpha kleq exp left(sum_ alpha kp k-1 right) beginprod_ alpha kleq exp left(sum_ alpha kp k-1 right) beginprod_ alpha kleq exp left(sum_ alpha kp k-1 right) beginprod_ alpha kle PRODUCTION LEVEL RIGHT (FRACTURE) LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL LEVEL So we have a 1p1, a 2p2, a 3p3, +anpnA=AA=1, sum_ fracp k = frac= frac= frac= 1, and it follows that exp(k=1nakApk1) = 1.exp left(sum fracp k-1right) = 1.

- In this case, we are driven to the conclusion that k=1n(akA)pk1, k=1nakpk1, k=1nApk1, etc., etc., etc., etc., etc., etc.
- Let SS represent a collection of positive real numbers.
- (a 1+a 2+cdots+a n) = underbrace The existence of a word that is the sum of all the terms of SS is well known to us at this point.
- The fact that every term is positive means that anynthntextpower expansion is bigger than its summand, and as a result, we obtain (i=1nai)Nn!
- It is written as left(sum_ ai right) and left(prod_ ai right) in the following format: Now all we have to do is recall the definition of geometric mean, which is the thenthntextroot of the product between all terms on a set, followed by some algebraic operations to arrive at the result.
- Now, we can establish by induction that for any positive integernn,nn!ngeq n!, and this proof will assist us in eliminating the factorial on the denominator of the previous inequality, which was previously mentioned.
- is formed as 11!1geq 1!, which is correct.

Then, here’s what we have: (k+1)k+1=kk+1+(k+1)⋅kk⋅11+k⋅(k+1)2⋅kk−1⋅12+⋯+1k+1kk+1+(k+1)⋅kkkk⋅(2k+1).

(2k+1)≥k!.

.kk⋅(2k+1)≥k⋅k!+(k+1)!(k+1)!

.beginkgeq k!

(k+1)!

Rightarrow k cdot(2k+1)(k+1)!

With(k+1)k+1 kk(2k+1)(k+1)kcdot (2k+1)andkk(2k+1)(k+1)!

kcdot (2k+1) (k+1)!, we get(k+1)k+1(k+1)!(k+1)k+1(k+1)!(k+1)k+1(k+1)!(k+1)k+1(k+1)!

However, fork=0k=0 In the previous section, we observed that there is equality, and it becomes (k+1)k+1(k+1)!(k+1)geq (k+1)!, as we sought to demonstrate.

1nn≤1n!

fracleq fracleq fracleq fracleq fracleq fracleq fracleq fracleq fracleq fracleq fracleq fracleq fracleq fracleq fracleq fracleq fracleq fracleq fracleq fracleq fracleq fracleq fracleq fracleq fracleq frac In this case, the arithmetic mean is mam a Now, using a geometric understanding of the real line, the following findings are simple to comprehend: kmg0⟹mg∈(0,k)k≥ma0⟹ma∈(0,k].

- m a 0 implies m a in (0, k).
- Otherwise, they are on an equal footing.
- square mgma.m g m a.
- square mgma.m g m a.
- square Application for AM-GM (Main Article) More information on the applications of AM-GM inequality may be found in the paper referenced above, as well as in the section below.
- The minimal value of (1+1x)(1+1y)left(1+fracright)left(1+fracright) is found if and only if x,y=8x,y=8 and x,y=8x,y=8.
- From AM-GMx+y2xy4xy16xy116xy1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1xy+1 As a result, the smallest possible value of(1+1x)(1+1y)left(1+fracright)left(1+fracright)left(1+fracright)left(1+fracright)is2516frac.

It should be noted that the equality is true whenx=y=4.x=y=4 They may be calculated by noting that the smallest value is attained whenxy=16,xy=16, together with the specified restriction thatx+y=8, and then multiplying that number by eight.

In this case, we use the 3-variable version of AM-GM withx1=a3, x2=a3, x 1=a3, and x 2 = a3.

Afterwards, we multiply both sides by three to get b2ba2 + a3 + b3geq 3a2ba3 + a3 + b3geq 3a2ba3 □_square Find all of the true answers to 2x+x2=212x2x + x2 = 2 – frac.2x+x2=2 – frac.2x+x2=2 – frac.2x+x2=2 – frac.2x+x2=2 – frac.2x+x2=2 – frac.2x+x2=2 – frac.2x+x2=2 – frac.

We have2x2+12x=2×22 leq 2x + frac= 2 – x2, so0x20 geq x2.

is the only value that can be used.

□_square Look for all of the true positive answers to 4x+18y=14,2y+9z=15,9z+16x=17.

end4x + frac=14,2y + frac=15,9z + frac=17.

Summation of the above three inequalities yields the following result: 416x, 2Y, 18Y, 9Z, 9Z16, 12 and 18=46.

4x + frac+ 2y + frac+ 9z + fracgeq 16 + 12 + 18 = 46.

Furthermore, by summing the three given equations, we obtain4x+16x+2y+18y+9z+9z=46.

With these values entered into the original equations, we observe that (x, y, and z)=(2, 3, 1)(x, y, and Z)=(2, 3, 1)(x, y, and Z) = (2, 3, 1)is in fact a solution.

Solution 1: By expanding both sides, we can cancel termsa2c2a2c2andb2d2b2d2, which means we must show thata2d2+b2c2acbda is equal to a2d2+b2c2acbda.

2d2 + B2 = 2aCBd.

2.big(a2+b2big)big(c2+d2big) = (ac+bd)2 + (ad-bc)2 = (ac+bd)2 + (ad-bc)2 = (ac+bd)2 + (ad-bc)2 = (ac+bd)2 + (ad-bc)2 = (ac+bd)2 + (ad-bc)2 = (ac The right-hand side is bigger than or equal to (ac+bd)2, which is non-negative since squares are non-negative.

abc(a+b+c) = a4 + b4 + c4 = abc(a+b+c).

Think about how we can acquire words on the right hand side of the screen by using AM-GM.

This provides a hint to trya4+a4+b4+c4a2bc.a4 + a4 + b4 + c4 eq 4 a2 b c.a4 + a4 + b4 + c4 eq 4 a2 b c.a4 + a4 + b4 + c4 eq 4 a2 b c.a4 + a4 + b4 + Similarly, we have a4+b4+b4+c4+4ab2ca4 +b4 +b4 +c4 geq 4ab2 c and a4+b4+c4+c4abc2.a4 + b4 + c4 + c4 geq 4abc2.a4 + b4 + c4 + c4 geq 4abc2.a4 + b4 + 2.

- In the case of abc(a+b+c), the equation is a4 + b4 + c4 = a2c + ab2c + ab2 = abc(a+b+c).
- frac+frac+frac+frac What is the smallest possible value ofP2A?frac for all rectangles with perimeterPP and areaAA?
- Examples of problems: Figure out what the smallest possible value of4+9x2sin2xxsinxfor0x.fractext 0xpi is.
- We can also write the expression 4+9x2SINX2XSINXfrac sin as 4xSINX.
- squareForx0x0, the maximum value is maximizef(x)=(1+x)(1+x)(1x), and the minimum value is 1212.
- Give your answer with as many decimal places as you choose.
- This time, we’ll apply the identical method to Jensen’s inequality that we used to show the AM-GM inequality above to demonstrate the weighted AM-GM inequality.
- It is true that allaka kare are equal if they are all equal.
- It follows that if at least one of the aka kis is zero (but not all), then the weighted geometric mean is zero, but the weighted arithmetic mean is positive, and that the stringent inequality applies in this case.

sayln(ω1a1+⋯+ωnanω)w1wlna1+⋯+wnwlnan=ln(a1ω1×2ω2⋯anωn∑ i).beginlnBigl(fracomega Bigr)frac wln a 1+cdots+frac wln a n =ln left(sqrtx 2cdots a n)ln right(sqrtx 2cdots a n)ln right(sqrtx 2cdots a

## Art of Problem Solving

Known technically as theInequality of Arithmetic and Geometric Meansor colloquially as AM-GM in mathematics, theAM-GM Inequality says that the arithmetic mean of any list of nonnegative reals is either greater than or equal to the geometric mean of that list. Furthermore, the two means are identical if and only if every number in the list is the same for each of the two groups of numbers. In symbols, the inequality asserts that for any real numbers, withequalityif and only if is true for any real numbers.

Applications may be found at the beginner, intermediate, and olympiad levels of problems, with AM-GM being particularly important in proof-based contests, among other places.

## Proofs

Induction and other more sophisticated inequalities are used in all known proofs of AM-GM, as described in the main article. They are also all more sophisticated than their application in the majority of beginning and intermediate contests, which is a good thing. AM-most GM’s simple proof makes use ofCauchy Induction, a type of induction in which one proves a result for, then uses induction to extend this to all powers of, and finally argues that assuming the result forimplies that it holds for all other powers of.

## Generalizations

It has proven possible to generalize the AM-GM Inequality into a number of different inequalities. Other generalizations of AM-GM include theMinkowski Inequality andMuirhead’s Inequality, which are in addition to those already mentioned.

### Weighted AM-GM Inequality

The Weighted Arithmetic-Geometric Means Inequality is a relationship between the weighted arithmetic and geometric means According to the rule, given any list of weights such that, with equality if and only if When the weighted form is reduced to the AM-GM Inequality, the result is a positive value. The proofs of the AM-GMarticle contain a number of proofs of the Weighted AM-GM Inequality.

### Mean Inequality Chain

The Mean Inequality Chain is the main article. The Root Mean Square, Arithmetic Mean, Geometric Mean, and Harmonic Mean of a list of nonnegative reals are all related by the Root Mean Square, Arithmetic Mean, Geometric Mean, and Harmonic Mean Inequality Chain. It specifically specifies that with equality if and only if the following conditions are met The Mean Inequality Chain, like the AM-GM, has a weighted variant that is similar to the AM-GM.

### Power Mean Inequality

Power Mean Inequality is the main article. The Power Mean Inequality is a mathematical relationship between all of the distinct power means of a list of nonnegative real numbers. The following is how the power mean is defined: It follows from this that ifb$” width=”42″ height=”13″> and equality holds only ifb$” width=”42″ height=”13″>, then plugginginto this inequality reduces it to AM-GM, and produces the Mean Inequality Chain. A weighted version of the Power Mean Inequality exists, just as there is a weighted version of the AM-GM.

## Problems

- Inequality
- Mean Inequality Chains
- Power Mean Inequality
- Cauchy-Schwarz Inequality
- Proofs of AM-GM

## Geometric Mean vs Arithmetic Mean

In the case of a product, the geometric mean is calculated as the mean or average of a series of values that takes into account the effect of compounding and is used to determine the performance of the investment, whereas the arithmetic mean is calculated as the mean as the sum of the sum of the total of values divided by the number of values in the series of values. You are allowed to use this image on your website, in templates, or in any other way you see fit. Please credit us by include a link to this page.

As an illustration: Geometric Mean vs.

The Arithmetic Mean is simply the average, and it is computed by putting all of the numbers together and dividing the total number of numbers by the total number of numbers in that series.

### Geometric Mean vs. Arithmetic Mean Infographics

You are allowed to use this image on your website, in templates, or in any other way you see fit. Please credit us by include a link to this page. Hyperlinking an article link will be implemented. Source: Geometric Mean vs Arithmetic Mean, for instance (wallstreetmojo.com) You are allowed to use this image on your website, in templates, or in any other way you see fit. Please credit us by include a link to this page. Hyperlinking an article link will be implemented. Source: Geometric Mean vs Arithmetic Mean, for instance (wallstreetmojo.com)

### Key Differences

- The arithmetic mean, also known as the additive mean, is employed in the computation of returns on a daily basis. Geometric Mean is also known as multiplicative mean, and it is a bit more hard to calculate since it requires compounding. The fundamental difference between the two means is the method by which they are computed. Calculating the arithmetic mean The arithmetic mean is the average of all of the observations in a data series, and it is defined as It is defined as the sum of all the values in a data collection divided by the total number of observations in the data set. click here to find out more This is obtained by dividing the total number of data points by the number of numbers in the dataset. When a series of numbers is determined by taking the product of these numbers and rising it to the inverse of the length of the series, this is known as the geometric mean. Return1 + Return2 + Return3 + Return4/ 4 is the formula for the geometric mean, whereas the formula for the arithmetic mean is 1
- Geometric mean (Geometric mean) It is a sort of mean that employs the product of values often ascribed to a collection of numbers to reflect the usual values or central tendency of a group of numbers. When there is an exponential change in values, this approach may be used to calculate the change. click here to find out more The geometric mean can only be computed for positive integers and is always smaller than the arithmetic mean
- On the other hand, the arithmetic mean can be calculated for both positive and negative values and is always bigger than the geometric mean The impact of outliers on a dataset is one of the most typical issues that arise when dealing with large datasets. The geometric mean in a dataset of 11, 13, 17, and 1000 is 39.5, but the arithmetic mean is 260.75 in the same dataset. The significance of the impact is clearly demonstrated. Normalization of the dataset and averaging of values are achieved as a result of the geometric mean
- As a result, no range dominates the weights and no percentage has a substantial impact on the data set. Unlike the arithmetic average, which is affected by skewed distributions, the geometric mean is not affected by them
- The arithmetic mean is employed by statisticians, but only for data sets that do not contain any notable outliers. When it comes to reading temperatures, this form of mean is really beneficial. Moreover, it is beneficial in determining the average speed of the vehicle. When a dataset is logarithmic or changes by multiples of ten, the geometric mean, on the other hand, is beneficial. Many scientists use this form of mean to characterize the size of bacteria populations since it is easy to calculate and understand. For example, the bacterial population can fluctuate between 10 and 10,000 in a single day. It is also possible to compute the distribution of income using a geometric average formula. For example, X and Y each earn $30,000 per year, but Z earns $300,000 per year. In this situation, the arithmetic average will be of no assistance. Portfolio managers draw attention to A portfolio manager is a financial market specialist who is responsible for the construction of investment portfolios on a strategic basis. a description of how and by how much a person’s wealth has risen or diminished
- Read more

### Comparative Table

Basis | Geometric Mean | Arithmetic Mean |
---|---|---|

Meaning | Geometric Mean is known as the Multiplicative Mean. | Arithmetic Mean is known as Additive Mean. |

Formula | – 1 | (Return1 + Return2 + Return3 + Return4)/ 4 |

Values | The geometric mean is always lower than the arithmetic means due to the compounding effect. | The arithmetic mean is always higher than the geometric mean as it is calculated as a simple average. |

Calculation | Suppose a dataset has the following numbers – 50, 75, 100. Geometric mean is calculated as cube root of (50 x 75 x 100) = 72.1 | Similarly, for a dataset of 50, 75, and 100, arithmetic mean is calculated as (50+75+100)/3 = 75 |

Dataset | It is applicable only to only a positive set of numbers. | It can be calculated with both positive and negative sets of numbers. |

Usefulness | Geometric mean can be more useful when the dataset is logarithmic. The difference between the two values is the length. | This method is more appropriate when calculating the mean value of the outputs of a set ofindependent eventsIndependent event refers to the set of two events in which the occurrence of one of the events doesn’t impact the occurrence of another event of the set.read more. |

Effect of Outlier | The effect of outliers on the Geometric mean is mild. Consider the dataset 11,13,17 and 1000. In this case, 1000 is the outlier. Here, the average is 39.5 | The arithmetic mean has a severe effect of outliers. In the dataset 11,13,17 and 1000, the average is 260.25 |

Uses | The geometric mean is used by biologists, economists, and also majorly byfinancial analysts. It is most appropriate for a dataset that exhibits correlation. | The arithmetic mean is used to represent average temperature as well as for car speed. |

### Conclusion

If you are dealing with percentage changes, volatile figures or data that exhibits correlation, the geometric mean is the best option. This is especially true for investment portfolios. If you make a portfolio investment instead of a single asset, you are investing in a group of assets (equity or debt), with the goal of earning returns that are proportional to the investor’s risk profile. Portfolio investments can include anything from stocks to bonds to mutual funds to derivatives to bitcoins.

The majority of financial returns, such as stock returns, bond yields, and premiums, are connected.

While arithmetic means is more suited for independent data sets because it is straightforward to use and comprehend, it is less appropriate for dependent data sets.

### Recommended Articles

This article served as a comparison of the Geometric Mean and the Arithmetic Mean. With the use of infographics and a comparison table, we’ll go through the top nine distinctions between Geometric Mean and Arithmetic Mean in this article. Check out the following articles as well if you want to learn more:

- The power of compounding
- The formula for the weighted mean
- The difference between the mean and the median
- The annualized rate of return

Interactions with the reader

## The Difference Between the Arithmetic Mean and Geometric Mean

There are several methods for evaluating the performance of a financial portfolio and determining whether or not an investment plan is effective. The geometric average, often known as the geometric mean, is frequently used by investment experts to make decisions.

### Key Takeaways:

- In the case of series that display serial correlation, the geometric mean is the most appropriate choice. Specifically, this is true for investment portfolios because the majority of financial returns are connected, such as bond yields, stock returns, and market risk premiums, among other things. Compounding becomes increasingly crucial with increasing time horizon, and the usage of the geometric mean becomes more acceptable. Because it takes into account year-over-year compounding, the geometric average gives a significantly more accurate representation of the underlying return for volatile values.

Due to the compounding that happens from period to period, the geometric mean differs from thearithmetic mean, or arithmetic mean, in how it is determined. Investors generally believe that the geometric mean is a more accurate gauge of returns than the arithmetic mean as a result of this phenomenon.

## The Formula for Arithmetic Average

The portfolio returns for periodnn are represented by the numbers a1, a2,., ann where: a1, a2,., ann=Portfolio returns for periodnn=Number of periods begin A = fracsum_ n a i = fractextbfa 1, a 2, dotso, a n=textn n=textn n=textn n=textn n=textn n=textn n=textn n=textn n=textn n=textn n=textn n=textn n=textn n=textn n=textn n=text A=n1 i=1n ai =na1 +a2 +.+an where:a1,a2,.,an =Portfolio returns for periodnn=Number of periodsnn=Number of periodsnn=Number of periodsnn=Number of periodsnn=Number of periods

## How to Calculate the Arithmetic Average

An arithmetic average is the product of the sum of a series of numbers divided by the number of numbers in that series. If you were asked to calculate the class (arithmetic) average of test results, you would simply add up all of the students’ test scores and divide that total by the number of students in the class. For example, if five students completed an exam and had scores of 60 percent, 70 percent, 80 percent, 90 percent, and 100 percent, the average for the arithmetic class would be 80 percent.

- It is because each score is an independent event that we use an arithmetic average to calculate test scores instead of a simple average.
- It is not uncommon in the field of finance to find that the arithmetic mean is not an acceptable way for determining an average.
- Consider the following scenario: you have been investing your funds in the financial markets for five years.
- With the arithmetic average, the average return would be 12 percent, which looks to be a substantial amount at first glance—but it is not totally correct in this case.
- They are interdependent.

Our goal is to arrive at an accurate calculation of your actual average yearly return over a five-year period. To do so, we must compute the geometric average of your investment returns.

## The Formula for Geometric Average

x1,x2,=Portfolio returns for each periodn=Number of periodsbeginleft(prod_ n x i ) = sqrttextbfx 1, x 2, the number of periods is equal to the sum of the returns on the portfolio for each period (x1, x2,.) and the number of periods is equal to the sum of the returns on the portfolio for each period (n1).

## How to Calculate the Geometric Average

It is possible to determine the geometric mean for a series of integers by multiplying the product of these values by the inverse of the length of the series. In order to do this, we add one to each of the numbers (to avoid any problems with negative percentages). In the next step, add up all of the numbers and elevate their product to a power of one divided by the number of numbers in the series. Then we take one away from the final result. When expressed in decimal form, the formula looks somewhat like this: [(1+R1)×(1+R2)×(1+R3)…×(1+Rn)] beginning with the number 1 and ending with the number 1 where R=Returnn=Count of the numbers in the series starting with the number 1 and ending with the number 1 – 1 textbftext= textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = textn = text 1 where: R = Returnn = Number of numbers in the series 1 where: R = Returnn = Number of numbers in the series Although the formula appears complicated, it is not as difficult as it appears on paper.

Using our previous example, we can calculate the geometric average as follows: The percentages of returns we received were 90 percent, 10 percent, 20 percent, 30 percent, and -90 percent, therefore we entered them into the calculation as: (1.91.11.21.30.1).

The figure obtained by applying the geometric average is far worse than the 12 percent arithmetic average we obtained previously, and regrettably, it is also the number that best depicts reality in this particular instance as well.

## Using the Arithmetic Mean-Geometric Mean Inequality in Problem Solving.

Some of the information in this article was used to construct a presentation for the Annual Meeting of the School Mathematics and Science Association, which was held in Birmingham on November 8, 2012. Version in PDF format The Arithmetic Mean-Geometric Mean Inequality (AM-GM Inquality) is a mathematical relationship that is fundamental to many mathematical concepts. It is a valuable tool for problem solving and for establishing connections with other branches of mathematics. The use of this concept in classroom mathematics should be greater than it is today.

This paper provides a broad introduction to the theorem as well as some background information and extensions, alternate demonstrations of the proof, and instances of situations that may be investigated utilizing the AM-GM Inquality.

In my examples, I will concentrate on the theorem for two positive numbers, but I will also discuss the extensions below and occasionally use the case for three positive numbers to illustrate the point.

The link raises the question of whether it is possible to generalize the evidence in accordance with the lines of argument presented by Courant and Robbins (1942).

GEOMETRIC DEMONSTRATION OF THE RMS-AM-GM-HM INEQUALITY For two positive values a and b, in particular, Using a more sophisticated degree of analysis (maybe a more basic approach?) They are all instances of Power Methods, which are defined as follows: where the power parameter has distinct values for each of the different means These are sometimes referred to as Generalized Means.

- For example, it is commonly known that the maximum area of a rectangle with a set perimeter is equal to the area of a square with the same perimeter.
- Comment: The word “minimum” should be replaced with the word “maximum” in the second statement below.
- 2a + 2b = 10 is the fixed perimeter of the circle.
- Let brerepresent the length of one side and 5 – b represent the length of the opposite side of the equation.

points on the blue curve with Area = 6.25 are always greater than points on the red curve (that is, the area of a rectangular prism will never be greater than 6.25), and the blue curve (a horizontal line representing an invariant) is tangent to the red curve when and only when (b = (5-a) = (b), that is, when b =2.5.

The other side has a value of 5 – x.

When a =b, the area is always smaller than the constant, and when a =b, the area is equal to the constant.

Here is an example of a challenge addressed to students on my website: Unlike the first problem, in which ab is a constant rather thana + b, the second problem is straightforward and arises directly from the AM-GM Inequality: In this part, I will restrict the investigation to the most straightforward scenario: The arithmetic mean and geometric mean of two positive integers are both equal to one another.

In my Problem Solving course, I want the students to discover at least 5 demonstrations or proofs, which I refer to as an EXPLORATION.

Seeing a variety of approaches to this relationship can aid in comprehending it, recognizing its significance, and recognizing its use as a problem-solving tool in the future.

## Some algebra

Construct a semicircle with a diameter equal to the sum of the two numbers a and b. The Arithmetic Mean of a and b will be used to get the radius. From the common endpoints of the segments of lengtha andb, construct a line perpendicular to the diameter of the circle. Create the red segment by intersecting this perpendicular with the semicircle and extending it from there. This segment will always have a length that is less than or equal to the radius of the circle, and it will always be equal if and only if a and b have the same length.

## Another Geometry Example

It is necessary to start with the identity: Sincewe havewith equality if and only ifa = b, we can proceed to the demonstration. Another Geometry Exemplification Given two tangent circles with radii a and b, solve for a. Using the two circles, construct an exterior tangent that is common to both circles, then connect the radii of each circle to the common tangent. Calculate the length indicatedc by?along the common tanget in terms ofa and b using the formula. A right triangle with legs of length?

- The segment along the common tangent has length twice the geometric mean of the two segments.
- Allow the CD to pass through the midway M of the AB equation.
- The lengths are all in the positive direction.
- We can only be certain that x + y = 2z with equality if M is the midway of the CD.

## Cost of Fencing a Field

An agriculturalist wishes to fence in 60,000 square feet of land in a rectangular plot along a straight highway, posing the following problem: The fence he intends to use along the highway will cost him $2 per foot, but the fence he intends to use on the other three sides will cost him $1 per foot When purchasing fences, how much of each type will be required to ensure that costs are kept as low as possible?

What is the bare minimum in terms of expenditure? Solution: (This is a common calculus issue, yet there is no need for calculus here!)

## Minimum of

The objective is to determine the smallest possible values for this function in the range or0x. The AM-GM Inequality must be applied to this equation, which necessitates a modification in the equation’s form. To begin, though, it is beneficial to examine a graph and understand the equation in some detail. There may be two minimum values in the range0x according to the graph on the right, which proposes two possibilities.

## Construct a Square with Same Area as a Given Rectangle

Construction of a square with a straightedge and a compass that has the same area as a specified rectangle is the objective of the task.

The area of the square will be banded, and the length of the side of the square will be the geometric mean of the area. Consider: The geometric mean of a and b is shown by the red section.

## Maximum Area of a Sector of a Circle With Fixed Perimeter

Issue: Use a straightedge and compass to make a square with the same area as a specified rectangle, and then solve the problem. The area will be banded, and the geometric mean will equal the length of the side of the square. Consider: Each segment in red represents the geometric mean of the two segments in blue and yellow respectively.

## Sector as Fraction of a Circle

When it comes to dealing with radian measure, some students (as well as teachers) are uncomfortable. For them, the following technique would be appropriate: In this case, letkbe the fraction of a circle represented by the thesector 0k1. Because the perimeter is a constant, we may describe it as twice the radius multiplied by the proportion of the circumference of the circle. The area of the sector in fraction notation can be expressed as either a function of r or a function of k, depending on whether the perimeter equation is substituted for r or for k in the sector equation.

The angle is a fraction of a degree less than 120 degrees.

## Sector in Radian Measure

It is interesting to note the similarity between this solution and the previous one, in which the square was the maximum area for rectangles with set perimeter.

## Inequalities Problems

To complete a right triangle, construct segments from the diameter’s ends in the manner shown.

## Maximumand Minimum of

Rewrite The value of the function is always less than or equal to.5 when x0 is taken into consideration, according to the arithmetic mean-geometric mean inequality. When x = 1, the value of the function is equal to.5. When applied to x0, a similar parameter leads to the discovery of the minimum of the function when x = -1. We must apply the Arithmetic Mean – Geometric Mean Inequality to – x and -1/x when x0 is positive since the inequality holds only for positive numbers. We’re aware of it. Always remember that this is for x0, thus -x and -1/x are both positive.

In order to do this, the value of the function is always greater than or equal to -0.5, and it is only equal to -0.5 when x equals -1.

## Maximum off(x) = (1-x)(1+x)(1+x)

It is necessary for us to know when the function reaches its maximum in an interval since we have three elements in the function. In order to benefit from the AM-GMInequality, we must ensure that the total of these elements equals a fixed amount of money. That is why we require 2x rather than x. We may obtain this information by doing the following: In the case of x inif, the function achieves its maximum value only if2(x) = 1 + x2 – 2x = 1 + x1 = 3x = 1 + x2 – 2x = 1 + x1 = 3x = 1 + x2 – 2x = 1 + x1 = 3x = 1 + x1 = 3x = 1 + x1 = 3x That is to say,

## MinimumSurface Area of a Can of Fixed Volume

The form of the can (e.g., short and fat vs. tall and slim) that is used to package a product in a given volume may be determined by different criteria such as tradition and presumptive client preferences, among other things, in the case of a fixed volume. Take, for example, the fact that all 12 oz. Coke cans are the same form – with a height of around 5 inches and a radius of approximately 1.25 inches – Why? What if the decision was made with the goal of reducing the amount of material utilized to create the can?

What is the connection between the radius and the height when attempting to decrease the surface area of a fixed volume of material?

The volume is set at a certain level. Surface area is a function of the parameters r and h. What is the very bare minimal S? (Please keep in mind that we have used the AM-GM Inequality for three positive values.)

## Maximum area of a Pen, fixed length of fencing, with Partitions Parallel to a Side

Consider the following scenario: you have 100 feet of fencing. You’re looking for a rectangular pen with a divider running parallel to one of the sides. The 100 feet of fence must encircle all four sides of the building as well as the partition. What is the form of the pen, and how much space does it have at its maximum? Demonstrate that the maximum area penis has the following dimensions: 25 ft. by 16.67 ft. with a maximum area of roughly 417.7 sq. ft.

### Prove the Maximum Area of a Triangle with Fixed Perimeter is Equilateral

Find the point on the graph of that is the most closely associated with the origin. (Click on the link to get to the debate) Given a triangle with one side measuring 9 units in length and the ratio of the other two sides being 40/41, find the length of the other side. Calculate the largest possible area. (Discussion may be found at the link) Calculate the maximum volume of a box created from a 25by 25 square sheet of cardboard by removing a tiny square from each corner and folding up the edges to produce a lidless box using the Arithmetic Mean-Geometric MeanInequality.

Clearly, the proof of3implies the proofs of the previous two.

On the right is a portion of a spreadsheet displaying the results of the calculation for.

The Excel file may be downloaded by clicking HERE.

## The Arithmetic-Geometric Mean Inequality

Assume thatxandyare non-negative real numbers that are not necessarily different from one another. It is stated in the well-known arithmetic-geometric mean inequality that_x=y with equality if and only if This also applies to the situation of non-negative integers in general: Once again, equality is only achieved if and only if all of the numbers are equal. It is the average, often known as the arithmetic mean, of the values on the left-hand side of this inequality that is being discussed here.

- In order to understand what the geometric mean is, I’ll first describe what it is.
- and |BC|, and a hypotenuse with length AC.
- and BC, and the length of the hypotenuse is AC.
- and BC, respectively.
- The height of ABC is shown by the line BD.
- The Pythagorean Theorem provides the following proof:

## Proof of the inequality

There are a variety of approaches that may be used to demonstrate the AM-GM disparity. However, my favorite proof is one that was found by the great mathematician George Pólya, who is one of the world’s most famous mathematicians. I will begin by demonstrating the following lemma: xex1 represents all of the real x. To begin, I will define the function F(x)=ex1-x as follows: It is important to note that F(1)=0 and F′(x)=ex1-1. Case x 1 consists of the following: It follows that if F(x) is greater than zero for any x 1, then in order for F(x) to approach zero at x=1, F(x) would need to be rising for some x 1.

- As a result, for any x 1, F(x) must be positive.
- When x 1, x-1 0 so ex1 1 and so F′(x) 0, so F(x) is increasing for all x 1.
- This is because F(1)=0.
- As a result, F(x 1) 0, F(x 1) 0, and F(x=1)=0, resulting in xex1 for every real x.
- To complete the proof of the inequality, let x1 be the sum of all the non-negative numbers in the range x1 through xn.

Then: This may be summarized as follows: Because x1 is the arithmetic mean of x2,., xn, the argument of the exponential function may be reduced to the following: Then Since e0=1, the following is true: Now I’ll demonstrate that this inequality transforms into an equality if and only if all of the xi are of the same value.

=xn=: This brings the proof to a close.

## Demonstration: Turning hard calculus problems into easy inequality problems

In order to get the maximum and/or lowest (local or global) values of a given function F(x), the conventional procedure is to solve F′(x)=0 for x and then obtain the extreme values of F(x) by putting those x into F(x) (x). However, there are situations when the conventional strategy may not work. It is possible that solving F′(x)=0 for x will not be feasible, or that there may be other limitations in the issue. To our advantage, taking derivatives is not the sole method of solving optimization issues.

- I’ll give you two illustrations.
- My previous posts on this site have discussed the Putnam family heirloom collection.
- In the problem description, it is stated:Find the largest value of x3-3x while keeping the restriction x4+3613×2 in mind.
- Even if we were to begin, where would we begin?
- One thing that could pique our curiosity is that the AM-GM inequality says that the x4+36 is “connected” to 12×2, which is oddly close to the 13×2 in the dilemma.
- However, if I deduct 12×2 from the total inequity, I get the following:Now we’re in the black.

Because perfect squares are always non-negative, we can disregard the left inequality and just take the square root, and the restriction now takes on a much more straightforward structure: When this is true, the values of x that fall between the two values of x for which the graph of y=x crosses the graph of y=x2-6 are the values of x in between.

As a result, the limitation is reduced even more to: f(x) has a local maximum value of 2 at x=-2, and from there, f(x) increases monotonically as x increases to -1, where f(x) has a local minimum value of -2, and then f(x) decreases monotonically as x increases to 1 where f(x) has a local maximum value of 2, and f(x) increases monotonically from x=3 to x=3, and f(3)=18, so The second problem is Problem 4 from the British Mathematical Olympiad, which took place in 1991.

For the positive real integers x, y, and z, determine the value of (x+y)(y+z) that is the smallest possible while satisfying the restriction (xyz)(xyz)=1.

We’ll try to find a way around the tiresome jumble of Lagrange multipliers, which would be the bad news in this case.

The strategy used in the previous problem was a straightforward computation.

First, I’m going to demonstrate one conceivable manner in which you may guess that the correct answer is 2, and then I’ll demonstrate that 2 is the correct answer by demonstrating that 2 is the correct answer.

If we switch x and z, the value of f(x,y,z) remains unchanged; that is, f(x,y,z)=f(z,y,x) for all x and z.

So if f(a,b,c) is the minimal value then so is f(c,b,a) (c,b,a).

With that assumption, the issue reduces to minimizing (y+a)² subject to a²y(y+2a)=1.

Now I shall prove this.

But examine the expansion of xz(x+y)(y+z), bearing in mind that xyz(x+y+z)=1: So (x+y)(y+z) 2 would indicate that: This violates the AM-GM inequality, hence (x+y)(y+z) 2 is false for any (x,y,z) meeting the restriction.

The reason that this indirect technique is important is that it turns out that it’s actually quite difficult to establish there must exist a point (x,y,z) fulfilling x-z and where f(x,y,z) takes its minimal value.

It was studying those old notes and recalling how nice it felt to find out that I could use a basic inequality argument to bypass that much harder difficulty that prompted me to create this post.

## Conclusion

When used wisely, inequalities may be an effective problem-solving tool for tackling complex problems. When it comes to addressing specific physics difficulties, I have found them to be really valuable since they may occasionally provide beautiful answers to problems that would otherwise be extremely time-consuming to tackle directly using calculus. As an illustration, it’s a good idea to keep an eye out for new tools that you can incorporate into your arsenal of mathematical problem-solving techniques: even if you never really need them and you can’t predict when one or another tool will come in handy, having more tools will mean being able to solve more problems and solving problems in more effective ways in the long run.

## Copyright stuff

Unless otherwise specified, all of the photos in this page are my own creations. The Mathematical Association of America owns the intellectual property rights to the Putnam test problems. On the website kskedlaya.org, you may view previous Putnam questions dating back to the 1985 test, as well as solutions dating back to 1995. The British Mathematical Olympiad Subtrust owns the intellectual property rights to the problems from the British Mathematical Olympiad. In the book The Mathematical Olympiad Handbook, which is an excellent book containing all BMO problems from 1965 through 1996 as well as solutions and, uniquely among problem-solving books, extremely detailed commentary on those solutions and guidance on how to understand them, the version of the BMO problem that I wrote about here is taken from that book.