The next pictorial proof starts with two nonnegative numbers for example, 3 and 4 and compares the following two averages:

[ ext{ arithmetic mean } ≡ frac{3 + 4}{2} = 3/5; label{4.10}]

[ ext{ geometric mean } ≡ sqrt{3 imes 4} ≈ 3.464. label{4.11}]

Try another pair of numbers for example, 1 and 2. The arithmetic mean is 1.5; the geometric mean is (sqrt{2} ≈ 1.414). For both pairs, the geometric mean is smaller than the arithmetic mean. This pattern is general; it is the famous arithmetic-mean–geometric-mean (AM–GM) inequality [18]:

[egin{aligned}

&underbrace{frac{a+b}{2}}_{ ext {AM }} geqslant underbrace{sqrt{a b}}_{ ext {GM }}

& ext { inequality requires that } a, b geqslant 0.

end{aligned}label{4.12}]

Problem 4.5 More numerical examples

Test the AM–GM inequality using varied numerical examples. What do you notice when a and b are close to each other? Can you formalize the pattern? (See also Problem 4.16.)

## Symbolic proof

The AM–GM inequality has a pictorial and a symbolic proof. The symbolic proof begins with ((a−b)^{2}) a surprising choice because the inequality contains (a + b) rather than (a − b). The second odd choice is to form ((a − b)^{2}). It is nonnegative, so (a^{2} − 2ab + b^{2} geqslant 0). Now magically decide to add (4ab) to both sides. The result is

[underbrace{a^{2} + 2ab + b^{2}}_{(a + b)^{2}} geqslant 4ab label{4.13}]

The left side is ((a + b)^{2}), so (a + b geqslant 2 sqrt{ab}) and

[frac{a + b}{2} geqslant sqrt{ab}label{4.14}]

Although each step is simple, the whole chain seems like magic and leaves the *why *mysterious. If the algebra had ended with ((a + b)/4 geqslant ab), it would not look obviously wrong. In contrast, a convincing proof would leave us feeling that the inequality cannot help but be true.

## Pictorial proof

This satisfaction is provided by a pictorial proof.

Question

What is pictorial, or geometric, about the geometric mean?

A geometric picture for the geometric mean starts with a right triangle. Lay it with its hypotenuse horizontal; then cut it with the altitude (x) into the light and dark subtriangles. The hypotenuse splits into two lengths (a) and (b), and the altitude (x) is their geometric mean (sqrt{ab}).

Question

Why is the altitude (x) equal to (sqrt{ab})?

To show that (x = sqrt{ab}), compare the small, dark triangle to the large, light triangle by rotating the small triangle and laying it on the large triangle. The two triangles are similar! Therefore, their aspect ratios (the ratio of the short to the long side) are identical. In symbols, (x/a = b/x): The altitude x is therefore the geometric mean (sqrt{ab}). The uncut right triangle represents the geometric-mean portion of the AM–GM inequality. The arithmetic mean ((a + b)/2) also has a picture, as one-half of the hypotenuse. Thus, the inequality claims that

[frac{ ext{ hypotenuse } }{2} geqslant ext{ altitude } label{4.15}]

Alas, this claim is not pictorially obvious.

Question

Can you find an alternative geometric interpretation of the arithmetic mean that makes the AM–GM inequality pictorially obvious?

The arithmetic mean is also the radius of a circle with diameter (a + b). Therefore, circumscribe a semicircle around the triangle, matching the circle’s diameter with the hypotenuse (a + b) (problem 4.7.). The altitude cannot exceed the radius; therefore,

[frac{a + b}{2} geqslant sqrt{ab} label{4.16}]

Furthermore, the two sides are equal only when the altitude of the triangle is also a radius of the semicircle namely when (a = b). The picture therefore contains the inequality and its equality condition in one easy- to-grasp object. (An alternative pictorial proof of the AM–GM inequality is developed in Problem 4.33.)

Multiple problems

**Problem 4.6 Circumscribing a circle around a triangle**

Here are a few examples showing a circle circumscribed around a triangle.

Draw a picture to show that the circle is uniquely determined by the triangle.

**Problem 4.7 Finding the right semicircle**

A triangle uniquely determines its circumscribing circle (Problem 4.6). However, the circle’s diameter might not align with a side of the triangle. Can a semicircle always be circumscribed around a right triangle while aligning the circle’s diameter along the hypotenuse?

**Problem 4.8 Geometric mean of three numbers**

For three nonnegative numbers, the AM–GM inequality is

[frac{a + b + c}{3} geqslant (abc)^{1/3}. label{4.17}]

Why is this inequality, in contrast to its two-number cousin, unlikely to have a geometric proof? (If you find a proof, let me know.)

## Applications

Arithmetic and geometric means have wide mathematical application. The first application is a problem more often solved with derivatives: Fold a fixed length of fence into a rectangle enclosing the largest garden.

Question

What shape of rectangle maximizes the area?

The problem involves two quantities: a perimeter that is fixed and an area to maximize. If the perimeter is related to the arithmetic mean and the area to the geometric mean, then the AM–GM inequality might help maximize the area. The perimeter (P = 2(a + b)) is four times the arithmetic mean, and the area (A = ab) is the square of the geometric mean. Therefore, from the AM–GM inequality,

[underbrace{frac{P}{4}}_{ ext{ AM }} geqslant underbrace{sqrt{A}}_{ ext{ GM }} label{4.18}]

with equality when (a = b). The left side is fixed by the amount of fence. Thus the right side, which varies depending on (a) and (b), has a maximum of (P/4) when (a = b). The maximal-area rectangle is a square.

Multiple problems

**Problem 4.9 Direct pictorial proof**

The AM–GM reasoning for the maximal rectangular garden is indirect pictorial reasoning. It is symbolic reasoning built upon the pictorial proof for the AM– GM inequality. Can you draw a picture to show directly that the square is the optimal shape?

**Problem 4.10 Three-part product**

Find the maximum value of (f(x) = x^{2}(1 − 2x)) for (x geqslant 0), without using calculus. Sketch (f(x)) to confirm your answer.

**Problem 4.11 Unrestricted maximal area**

If the garden need not be rectangular, what is the maximal-area shape?

**Problem 4.12 Volume maximization**

Build an open-topped box as follows: Start with a unit square, cut out four identical corners, and fold in the flaps.

The box has volume (V = x(1 − 2x)^{2}), where x is the side length of a corner cutout. What choice of x maximizes the volume of the box?

Here is a plausible analysis modeled on the analysis of the rectangular garden. Set (a = x), (b = 1 − 2x), and (c = 1 − 2x). Then abc is the volume V, and (V^{1/3} = ^{3}sqrt{abc}) is the geometric mean (Problem 4.8). Because the geometric mean never exceeds the arithmetic mean and because the two means are equal when (a = b = c), the maximum volume is attained when (x = 1 − 2x). Therefore, choosing (x = 1/3) should maximize the volume of the box.

Now show that this choice is wrong by graphing (V(x)) or setting (dV/dx = 0); explain what is wrong with the preceding reasoning; and make a correct version.

**Problem 4.13 Trigonometric minimum**

Find the minimum value of

[frac{9x^{2} sin^{2}x + 4}{x sin x} label{4.19}]

in the region x (in) (0, (pi)).

**Problem 4.14 Trigonometric maximum**

In the region t (in) [0, (pi/2)], maximize (sin 2 t) or, equivalently, 2 (sin t cos t).

The second application of arithmetic and geometric means is a modern, amazingly rapid method for computing (pi) [5, 6]. Ancient methods for computing π included calculating the perimeter of many-sided regular polygons and provided a few decimal places of accuracy.

Recent computations have used Leibniz’s arctangent series

Imagine that you want to compute π to (10^{9}) digits, perhaps to test the hardware of a new supercomputer or to study whether the digits of π are random (a theme in Carl Sagan’s novel *Contact *[40]). Setting (x = 1) in the Leibniz series produces (pi/4), but the series converges extremely slowly. Obtaining (10^{9}) digits requires roughly (10^{10^{9}}) terms—far more terms than atoms in the universe.

Fortunately, a surprising trigonometric identity due to John Machin (1686 – 1751) accelerates the convergence by reducing (x):

Even with the speedup, (10^{9})-digit accuracy requires calculating roughly (10^{9}) terms.

In contrast, the modern Brent–Salamin algorithm [3, 41], which relies on arithmetic and geometric means, converges to π extremely rapidly. The algorithm is closely related to amazingly accurate methods for calculating the perimeter of an ellipse (Problem 4.15) and also for calculating mutual inductance [23]. The algorithm generates several sequences by starting with (a_{0}) = 1 and (g_{0}) = 1/(sqrt{2}); it then computes successive arithmetic means (a_{n}), geometric means (g_{n}), and their squared differences (d_{n}).

[a_{n+1}=frac{a_{n}+g_{n}}{2}, quad g_{n+1}=sqrt{a_{n} g_{n}}, quad d_{n}=a_{n}^{2}-g_{n}^{2} label{4.23}]

The a and g sequences rapidly converge to a number M((a_{0}), (g_{0})) called the arithmetic–geometric mean of (a_{0}) and (g_{0}). Then M((a_{0}), (g_{0})) and the difference sequence d determine π.

[pi=frac{4 Mleft(a_{0}, g_{0} ight)^{2}}{1-sum_{j=1}^{infty} 2^{j+1} d_{j}} label{4.24}]

The d sequence approaches zero quadratically; in other words, (d_{n+1}) ∼ (d^{2}_{n}) (Problem 4.16). Therefore, each iteration in this computation of π doubles the digits of accuracy. A billion-digit calculation of π requires only about 30 iterations far fewer than the 10109 terms using the arctangent series with (x = 1) or even than the (10^{9}) terms using Machin’s speedup.

Multiple problems

**Problem 4.15 Perimeter of an ellipse**

To compute the perimeter of an ellipse with semimajor axis (a_{0}) and semiminor axis (g_{0}), compute the a, g, and d sequences and the common limit M((a_{0}, g_{0})) of the a and g sequences, as for the computation of (pi). Then the perimeter P can be computed with the following formula:

[P=frac{A}{Mleft(a_{0}, g_{0} ight)}left(a_{0}^{2}-B sum_{j=0}^{infty} 2^{j} d_{j} ight)label{4.25}]

where A and B are constants for you to determine. Use the method of easy cases (Chapter 2) to determine their values. (See [3] to check your values and for a proof of the completed formula.)

**Problem 4.16 Quadratic convergence**

Start with (a_{0} = 1) and (g_{0} = 1/sqrt{2}) (or any other positive pair) and follow several iterations of the AM–GM sequence

[a_{n + 1} = frac{a_{n}+ g_{n}}{2} ext{ and } g_{n + 1} = sqrt{a_{n}g_{n}}. label{4.26}]

Then generate (d_{n} = a^{2}n − g^{2}n) and (log_{10}) dn to check that (d_{n+1} ∼ d^{2}_{n}) (quadratic convergence).

**Problem 4.17 Rapidity of convergence**

Pick a positive (x_{0}); then generate a sequence by the iteration

[x_{n + 1} = frac{1}{2}(x_{n} + frac{2}{x_{n}}) (n geqslant 0) label{4.27}]

To what and how rapidly does the sequence converge? What if (x_{0} < 0)?

The Arithmetic mean and Geometric mean are the tools widely used to calculate the returns on investment for investment portfolios in the world of finance. People use the arithmetic mean to report the higher returns which are not the correct measure of calculating the return on investment. Since the return on investment for a portfolio over the years is dependent on returns in previous years, the Geometric mean is the correct way to calculate the return on investment for a specific time period. The arithmetic mean is better suited in the situation wherein variables being used for the calculation of average are not dependent on each other.

**Example: Suitability use of Geometric mean vs Arithmetic mean**

Download Corporate Valuation, Investment Banking, Accounting, CFA Calculator & others

1. Let’s take an example of return on investment for an amount of $100 over 2 years. Suppose the returns in two years were -50% and +50% in the 1 st and 2 nd Average return calculation by using arithmetic mean will be 0% **(Arithmetic mean = (-50%+50%) /2 = 0%)**

This gives a wrong impression that the investor is breaking even on its investment, and there is no loss or profit. However, a closer analysis gives an entirely different picture of the scenario.

From the above table, we can see that the investment of $100 after -50% and +50% return in year 1 and 2, will be close to $75.Therefore, the investor is not breaking even on its investment as suggested by the arithmetic mean average, but he has incurred a loss of $25 after 2 years on its investment. This is well reflected by using Geometric mean to calculate the return on the investment over 2 years as below:

The geometric mean of returns

This means the annualized return on the portfolio had been **negative** at 13.40%. The investment position after two years is as below:

Therefore, the Geometric mean shows the true picture of investment that there is a loss in investment with an annualized negative return of -13.40%. Since the return in each year impacts the absolute return in the next year, a geometric mean is a better way to calculate the annualized return on investment.

2. When one needs to calculate the average of variables that are not dependent on each other, Arithmetic means a suitable tool to calculate the average. The average of marks of a student for 5 subjects can be calculated by the arithmetic mean as scores of the student in different subjects are independent of each other.

## 4.2: Arithmetic and Geometric Means

There are several methods for measuring the central tendency of a set of numbers.

One method is to calculate the arithmetic mean . To do this, add up all the values and divide the sum by the number of values. For example, if there are a set of “n” numbers, add the numbers together for example: a + b + c + d and so on. Then divide the sum by “n”.

One problem with the arithmetic mean is that its value will be influenced disproportionately by a single extreme value.

Another method is to calculate the geometric mean . To do this, multiply the values together and then, if there were “n” numbers, take the “n th ” root. Single extreme values then have less influence.

This method is particularly useful when results are recorded in logarithmic notation. To multiply you only have to add the log indices. To approximate the geometric mean, you take the arithmetic mean of the log indices.

You have recorded the following set of values in a serological test. To calculate the arithmetic mean, you must transform these to real numbers.

Calculation of the geometric mean = 4 Ö(8 × 16 × 16 × 64) = 4 Ö(131072) = 19

The root mean square, which is also commonly referred to as the quadratic mean, is frequently employed in statistical and engineering applications, especially when negative data points are under consideration. One example of the root mean square is the standard deviation of a set of numbers (I.e., it is the root mean square of the variations between the arithmetic mean and each data point).

When given two numbers, *x* and *y*, the quadratic mean is sqrt[(x 2 + y 2 )/2] .

## Geometric Mean Calculator

Geometric Mean is a type of mean or average that indicates the central tendency or typical value of a set of given numbers.

Geometric Mean is defined as the nth root of the product of the n units in a data set.

Geometric mean is a kind of average of a set of numbers that is different from the arithmetic average. Geometric mean is calculated for sets of positive real numbers. This is calculated by multiplying all the numbers (call the number of numbers n), and taking the nth root of the total.

Geometric Mean is used in the case when finding an average for set of numbers presented as percentages.

The geometric mean is an average that is useful for sets of positive numbers that are interpreted according to their product and not their sum (as is the case with the arithmetic mean) e.g. rates of growth. The online Geometric Mean Calculator is useful in calculating the geometric mean for the given set of numbers.

__ Example__:

Calculate the geometric mean for the given set of numbers.

25,56,85,71,4,12,3,2,5

* Solution*:

**Apply Formula:**

Geometric Mean = ((X

_{1})(X

_{2})(X

_{3}). (X

_{N})) 1/N

N = 9

1/N = 1/9

1/N = 0.111

Geometric mean = [(25)(56)(85)(71)(4)(12)(3)(2)(5)] 0.111

**Total Numbers: 9**

## Geometric V. Arithmetic Mean

From Dave McGlasson, does anyone have a reference on why a geometric mean is used in the calculation? Since we are using 20–30 normal subjects to calculate the mean , the range is relatively narrow. The geometric mean indicates the central tendency or typical value of a set of numbers by using the product of their values as opposed to the arithmetic mean, which uses their sum. I have calculated several data sets side by side comparing both arithmetic and geometric means and have never seen a clinically significantly difference that would require a physician to change a treatment plan.

The is faulty above 4.5 and should be considered semiquantitative. The curve flattens out. When comparing different reagent and instrument combinations the varies above 3.0.

I advocate doing away with the and going to using the chromogenic factor X (CFX) assay which reports the direct factor level. There are several studies that propose ranges for the CFX that correspond with the 2.0–3.0 therapeutic range and we don't have to worry about a faulty formula.

Dave provides the figure of the left from *Rosborough TK, Jacobsen JM, Shepherd MF. Relationship between chromogenic factor X and differs during early Coumadin initiation compared with chronic warfarin administration. Blood Coagul Fibrinolysis 2009 20:433-5*. The figure illustrates the relationship between the CFX, X scale, and , Y scale. Notice that the and CFX diverge as the exceeds 3.0.

Likewise, the figure above, from *McGlasson DL, Romick BG, Rubal BJ. Comparison of a chromogenic factor X assay with for monitoring oral anticoagulation therapy. Blood Coag Fibrinolys 200819:513–17*, illustrates the irrelevance of values when compared to CFX results when the exceeds 3.0.

From Dave McGlasson, does anyone have a reference on why a geometric mean is used in the calculation? Since we are using 20–30 normal subjects to calculate the mean , the range is relatively narrow. The geometric mean indicates the central tendency or typical value of a set of numbers by using the product of their values as opposed to the arithmetic mean, which uses their sum. I have calculated several data sets side by side comparing both arithmetic and geometric means and have never seen a clinically significantly difference that would require a physician to change a treatment plan.

The is faulty above 4.5 and should be considered semiquantitative. The curve flattens out. When comparing different reagent and instrument combinations the varies above 3.0.

I advocate doing away with the and going to using the chromogenic factor X (CFX) assay which reports the direct factor level. There are several studies that propose ranges for the CFX that correspond with the 2.0–3.0 therapeutic range and we don't have to worry about a faulty formula.

Dave provides the figure of the left from *Rosborough TK, Jacobsen JM, Shepherd MF. Relationship between chromogenic factor X and differs during early Coumadin initiation compared with chronic warfarin administration. Blood Coagul Fibrinolysis 2009 20:433-5*. The figure illustrates the relationship between the CFX, X scale, and , Y scale. Notice that the and CFX diverge as the exceeds 3.0.

Likewise, the figure above, from *McGlasson DL, Romick BG, Rubal BJ. Comparison of a chromogenic factor X assay with for monitoring oral anticoagulation therapy. Blood Coag Fibrinolys 200819:513–17*, illustrates the irrelevance of values when compared to CFX results when the exceeds 3.0.

## Example usage in Finance

When you evaluate an offer for a deposit with compound interest, or the expected returns from an investment strategy, you need to use the geometric average, not the arithmetic average. Let's see a quick example: if you hold money in a mutual fund for two years and it increased the value of its shares by 10% on the first year, and lost 10% on the second year, by using the arithmetic average of (15% - 15%)/2 = 0% you would expect to be where you started, but in fact you would have lost 2.25% of your initial investment (1.15 x 0.85)^ 1 /_{2} = 0.9775 or 97.75%, losing an average of 1.13% per year.

For a more complex example, let's say you are evaluating a strategy that projects the following return on investment for the next 5 years: 6%, 7%, 8%, -35%, 10%. The arithmetic average would be 0.4% return, but the **actual average yearly return** over those 5 years would be -2.62%, thus it will lose you money, despite having a positive return in 4 out of 5 years.

Period | Starting capital | % growth | End capital |
---|---|---|---|

1-st year | $1,000 | 6% | $1,060 |

2-nd year | $1,060 | 7% | $1,134.2 |

3-rd year | $1,134.2 | 8% | $1,224.94 |

4-th year | $1,224.94 | -35% | $796.21 |

5-th year | $796.21 | 10% | $875.83 |

Using the arithmetic average of 0.4% growth per year we expect to see an end capital of $1020.16, with the geometric average of -2.62% we see exactly $875.83.

## Proofs

### 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 the rearrangement inequality. Considering length and required prerequisites, the proof by induction given below is probably the best recommendation for first reading.

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

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

*i*=ف.

*n*.

For the following proof we apply mathematical induction and only well-known rules of arithmetic.

**Induction basis:** For *n* =ف 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* +ف non-negative real numbers. Their arithmetic mean *μ* satisfies

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

Now consider the *n* numbers

with

which are also non-negative. Since

*μ* is also the arithmetic mean of and the induction hypothesis implies

in particular *μ* >ـ. 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 get

which completes the proof.

### Second proof by induction

The following proof uses mathematical induction and some basic 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 we need to prove the following inequality:

We define and consider the following function:

Proving the induction step is equivalent to showing that for every , . It can be done by showing that for every , the minimal value of is greater or equal to 0.

After small rearrangement we get:

It is easily seen that at , . It implies that is a local minimum. Next we compute and show that it is greater or equal to 0, regardless of

The final inequality holds because of induction hypothesis. Finally we need to check that :

Summarizing all of the above we get:

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*

### Proof by Pólya

George Pólya provided a proof similar to what follows. Let *f*(*x*) = *e* *x*−1 − *x*, with derivative *f* ' (*x*) = *e* *x*−1 − 1. Observe *f* ' (1) = 0 and hence that *f* has an absolute minimum of *f*(1) = 0. Now *x* ≤ e *x*−1 for all real *x*.

Consider a list of non-negative numbers with arithmetic mean *μ*. By repeated application of the above inequality, we obtain the following:

But the exponential argument can be simplified:

which produces the result: [ 2 ]

### 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:

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* >ف.

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:

##### The subcase where n = 2 k

Consider the case where *n* =ق *k* , where *k* is a positive integer. We proceed by mathematical induction.

In the base case, *k* =ف, so *n* =ق. We have already shown that the inequality holds where *n* =ق, so we are done.

Now, suppose that for a given *k* >ف, 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:

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

(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:

##### 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:

### Old Classical Proof

If then replacing both and with will leave the left hand side unchanged, but will increase the right hand side as

Thus right hand side will be largest when all s are equal to (say) , thus as this is the largest value of right hand side of the expression, we have

### Proof of the generalized AM–GM inequality using Jensen's inequality

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

*α*=ـ has no influence on the inequality, we may assume in the following that all weights are positive. If all

_{k}*x*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

_{k}*x*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

_{k}*x*are positive.

_{k}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,

## Geometric Mean vs Arithmetic Mean

Geometric mean is the calculation of mean or average of series of values of product which takes into account the effect of compounding and it is used for determining the performance of investment whereas arithmetic mean is the calculation of mean by sum of total of values divided by number of values.

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The geometric mean is calculated for a series of numbers by taking the product of these numbers and raising it to the inverse length of the series. Arithmetic Mean is simply the average and is calculated by adding all the numbers and divided by the count of that series of numbers.

### Geometric Mean vs. Arithmetic Mean Infographics

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### Key Differences

- The arithmetic mean is known as additive mean and are used in the everyday calculation of returns. Geometric Mean is known as multiplicative mean and is a little complicated and involves compounding.
- The main difference in both these means is the way it is calculated. The arithmetic meanArithmetic MeanArithmetic mean denotes the average of all the observations of a data series. It is the aggregate of all the values in a data set divided by the total count of the observations.read more is calculated as the sum of all the numbers divided by the number of the dataset. The geometric mean is a series of numbers calculated by taking the product of these numbers and raising it to the inverse of the length of the series.
- Formula for geometric mean is <[(1+Return1) x (1+Return2) x (1+Return3)…)]^(1/n)]>– 1 and for arithmetic mean is (Return1 + Return2 + Return3 + Return4)/ 4. can only be calculated for positive numbers and is always less than geometric meanwhile, arithmetic mean can be calculated for both positive and negative numbers and is always greater than the geometric mean.
- A most common problem with having a dataset is the effect of outliers. In a dataset of 11, 13, 17, and 1000 the geometric mean is 39.5, while the arithmetic means 260.75. The effect is clearly highlighted. Geometric mean normalizes the dataset, and the values are averaged out hence, no range dominates the weights, and any percentage does not significantly affect the data set. The geometric mean is not influenced by skewed distributions as the arithmetic average is.
- The arithmetic mean is used by statisticians but for data set with no significant outliers. This type of mean is useful for reading temperatures. It is also helpful in determining the average speed of the car. On the other hand, the geometric mean is useful in cases where the dataset is logarithmic or varies by multiples of 10.
- Many biologists use this type of mean to describe the size of the bacterial population. For example, the bacterial population can be 10 in one day and 10,000 on others. Income distribution can also be calculated using a geometric average. For example, X and Y make $30,000 yearly, while Z makes $300,000 annually. In this case, the arithmetic average will not be useful. Portfolio managers highlightPortfolio Managers HighlightA Portfolio Manager is an executive responsible for making investment decisions & handle investment portfolios for fulfilling the client’s investment-related objectives. Also, he/she works towards maximizing the benefits & minimizing the potential risks for clients.read more how the wealth and by how much wealth of an individual has increased or decreased.

### 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) x (1+Return2) x (1+Return3)…)]^(1/n)]>– 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 of independent events. |

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 by financial 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

The use of geometric mean is appropriate for percentage changes, volatile numbers, and data that exhibit correlation, especially for investment portfolios. Most returns in finance are correlated like stocks, the yield on bonds, and premiums. The longer period makes the effect of compounding more critical and hence also the use of a geometric mean. While for independent data sets, arithmetic means is more appropriate as it is simple to use and easy to understand.

### Recommended Articles

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## Progressions - Arithmetic mean and Geometric mean

- As b - a = c - b => 2b = a + c or . Here 'b' is called the Arithmetic mean between 'a' and 'c'. In general, Arithmetic mean of the 'n' terms is equal to their average.

###### Geometric Progression

- Three numbers a, b and c are said to be in Geometric progression if i.e. if the ratio of the terms is same. This ratio of the terms is called the common ratio.
- Eg. 4, 16, 64, 256, 1024&hellip.. is a Geometric progression as the ratio of the terms is same.

- As => b 2 = ac or . Here 'b' is called the Geometric mean between 'a' and 'c'. In general, the geometric mean of 'n' numbers x
_{1}, x_{2}, x_{3}. x_{n}is given by