Means and Inequalities based on them

Introduction

We are going to learn three fundamental quantities and inequalities based on them. Trivial as they may seem in the beginning but during this course we will realize their importance as we will using them frequently thoughout.

Arithmetic Mean

The arithmetic mean \(A\) of two numbers \(a\) and \(b\) can be defined as a number lying between \(a\) and \(b\) such that \(a\), \(A\), and \(b\) are in Arithmetic Progression (AP). This implies that the difference between \(A\) and \(a\) is equal to the difference between \(b\) and \(A\), leading to the equation \(A - a = b - A\). Solving this equation for \(A\) gives:

\[ A = \frac{a + b}{2} \]

On a number line \(A\) bisects the line segment joining \(a\) and \(b\)

Arithmetic Mean by itself is not so intriguing or useful. It's usefulness is apprent when we compare it with Geometric Mean and Harmonic Mean.

Geometric Mean

The geometric mean of two positive numbers, \(a\) and \(b\), is defined as a positive number \(g\) that lies between \(a\) and \(b\) such that \(a\), \(g\), and \(b\) form a geometric progression (GP). This means that the ratio of \(g\) to \(a\) is the same as the ratio of \(b\) to \(g\). In mathematical terms, this relationship is expressed as:

\[ \frac{g}{a} = \frac{b}{g} \]

To find the value of \(g\), we can rearrange this equation and solve for \(g\). Multiplying both sides by \(g\) and then by \(a\) gives us:

\[ g^2 = ab \]

Taking the square root of both sides to solve for \(g\) yields:

\[ g = \sqrt{ab} \]

Geometric Mean of Non-positive numbers

The geometric mean, in its most common definition, is primarily used for sets of positive numbers. When the set includes non-positive numbers (i.e., zero or negative numbers), the concept of geometric mean encounters some complications:

Zero: If either of the two numbers is zero, the product of the two numbers will also be zero. The geometric mean, in this case, would be the square root of zero, which is zero. This scenario is mathematically sound but often not particularly useful in practical applications where the geometric mean is expected to represent a kind of "average" value.
Negative Numbers: For two negative numbers, their product will be positive, and thus, it is possible to calculate the geometric mean as the square root of this positive product. This scenario can be meaningful in certain mathematical or theoretical contexts but is less common in typical applications of the geometric mean, which often assume positive values for reasons of interpretability and relevance.
Mixed Numbers: When one number is negative and the other is positive, their product is negative, making the square root of their product undefined in the real numbers. This situation illustrates why the geometric mean is not typically defined for sets that include negative numbers when the context requires a real-number result.

In summary, while the geometric mean of two positive numbers is a well-defined and widely used concept, extending this definition to include non-positive numbers introduces complications. For zero, it's straightforward but less useful, and for mixed positive and negative numbers, it's mathematically infeasible within the real numbers. For two negative numbers, while feasible, it's less commonly applied in the traditional contexts where geometric means are used.

Harmonic Mean

The harmonic mean of two numbers, \(a\) and \(b\), is defined as a number \(h\) such that \(a\), \(h\), and \(b\) form a harmonic progression (HP). In a harmonic progression, the reciprocals of the terms form an arithmetic progression (AP). This means that if \(a\), \(h\), and \(b\) are in HP, then \(\frac{1}{a}\), \(\frac{1}{h}\), and \(\frac{1}{b}\) are in AP.

For three numbers to be in AP, the middle term must be the average of the other two. Therefore, the condition for \(a\), \(h\), and \(b\) to be in HP is:

\[ \frac{1}{h} = \frac{\frac{1}{a} + \frac{1}{b}}{2} \]

Solving this equation for \(h\) gives us the harmonic mean:

\[ \frac{1}{h} = \frac{1}{2} \left( \frac{1}{a} + \frac{1}{b} \right) \]

\[ 1/h = \frac{1}{2} \cdot \frac{2}{a+b} \cdot a \cdot b \]

\[ h = \frac{2}{\frac{1}{a} + \frac{1}{b}} = \frac{2ab}{a+b} \]

So, the harmonic mean \(h\) of two numbers \(a\) and \(b\) is \(\frac{2ab}{a+b}\).

Harmonic Mean does not always lie between \(a\) and \(b\)

When both \(a\) and \(b\) are positive or both are negative, the harmonic mean \(h = \frac{2ab}{a + b}\) indeed lies between \(a\) and \(b\). This is because the arithmetic involving positive or negative numbers (exclusively in each case) ensures that \(h\) is calculated as a value that respects the natural ordering of \(a\) and \(b\). In the case of both numbers being positive, \(h\) is a positive value less than the larger number and greater than the smaller number, due to the averaging effect of the harmonic mean formula. Similarly, when both numbers are negative, \(h\) remains between them because the arithmetic operations involved still respect the ordering of the numbers on the negative side of the number line.

On the contrary, when \(a\) and \(b\) have different signs, the harmonic mean \(h\) does not lie between \(a\) and \(b\).

For example,

For \(a = -1\) and \(b = \frac{1}{3}\), the calculation gives a harmonic mean (HM) of \(1\). This result highlights an important consideration: when \(a\) and \(b\) have opposite signs, the interpretation of the harmonic mean's position relative to \(a\) and \(b\) can be counterintuitive or not applicable in the typical sense of lying "between" the two values, since one is negative and the other is positive.

There is another problem. HM does not exist when \(a+b=0\), because in the formula as you can see this will result in division by zero.

Relationship between Arithmetic Mean, Geometric Mean and Harmonic Mean

For two positive numbers \(a\) and \(b\), let \(A\), \(G\), and \(H\) represent their Arithmetic Mean (AM), Geometric Mean (GM), and Harmonic Mean (HM) respectively. The formulas for these means are given by:

\[A = \frac{a + b}{2}\]

\[G = \sqrt{ab}\]

\[H = \frac{2ab}{a + b}\]

There exists a simple relationship between them.

\[G^2 = AH\]

To prove this proceed as follows:

Substitute the expressions for \(A\) and \(H\) into the right side of the desired equation:

\[AH = \left(\frac{a + b}{2}\right) \left(\frac{2ab}{a + b}\right)\]

Notice that the \((a + b)\) terms in the numerator and the denominator cancel out, simplifying to:

\[AH = ab\]

Now, consider the expression for \(G^2\):

\[G^2 = (\sqrt{ab})^2 = ab\]

Therefore, we have shown that:

\[G^2 = ab = AH\]

This relationship suggests that the geometric mean \(G\) of two positive numbers \(a\) and \(b\) is also the geometric mean of their arithmetic mean \(A\) and harmonic mean \(H\).

The fundamental inequality of AM, GM and HM

Theorem: For any two positive numbers \(a\) and \(b\), the following inequality holds: \(AM \geq GM \geq HM\), where \(AM\), \(GM\), and \(HM\) are the Arithmetic Mean, Geometric Mean, and Harmonic Mean of \(a\) and \(b\) respectively.

Also, \(AM=GM=HM\) if and only if \(a=b\)

Proof:

Let's start by establishing the relationship between the Arithmetic Mean (AM) and the Geometric Mean (GM).

Given two positive numbers \(a\) and \(b\), their AM and GM are defined as:

\(AM = \frac{a + b}{2}\)
\(GM = \sqrt{ab}\)

To prove \(AM \geq GM\), we utilize the identity:

\[ (a+b)^2 = (a-b)^2 + 4ab \]

We know that \((a-b)^2 \geq 0\). Hence, we can infer that:

\[ 4ab \leq (a+b)^2 \]

Dividing both sides by \(4\) gives:

\[ ab \leq \left(\frac{a+b}{2}\right)^2 \]

Taking the square root of both sides, and noting that \(a, b > 0\), we have:

\[ \sqrt{ab} \leq \frac{a+b}{2} \]

Therefore:

\[ GM \leq AM \]

Equality holds, that is, \(AM=GM\) if and only if \(a = b\). We started the derivation from the observation \((a-b)^2\ge 0\). It can be observed that equality holds, that is, \((a-b)^2=0\) only when \(a=b\). This equlaity is propagated to in the whole proof.

To prove \(GM \geq HM\), we start from the previously proven inequality \(AM \geq GM\) and rearrange it to derive an inequality involving \(HM\).

Given that:

\[ \frac{a + b}{2} \geq \sqrt{ab} \]

Taking the reciprocal of both sides (noting that reciprocals invert the inequality for positive numbers) gives:

\[ \frac{2}{a + b} \leq \frac{1}{\sqrt{ab}} \]

Multiplying both sides by \(2ab\) (positive, thus preserving the inequality direction) yields:

\[ \frac{2ab}{a + b} \leq \sqrt{ab} \]

Recognizing the left side as \(HM\) and the right side as \(GM\), we have:

\[ HM \leq GM \]

This demonstrates that for two positive numbers \(a\) and \(b\), the Geometric Mean is greater than or equal to the Harmonic Mean.

Conclusion: The ordered relationship \(AM \geq GM \geq HM\) is established for any two positive numbers \(a\) and \(b\), with equality in all means occurring if and only if \(a = b\). This completes the proof, illustrating the fundamental relationship between the arithmetic, geometric, and harmonic means.

Application of \(AM \ge GM\)

We will be discussing applications of this theorem in maxima and minima chapter of differential calculus. We will be also discussing in detail usage of this fundamental inequality and other inequalities in Inqualities chapter.

Inserting n Arithmetic Means (AM's) Between Two Numbers

When we talk about inserting \(n\) Arithmetic Means (AM's) between two numbers \(a\) and \(b\), we are referring to the process of finding \(n\) numbers \(a_1, a_2, a_3, \ldots, a_n\) such that the entire sequence \(a, a_1, a_2, a_3, \ldots, a_n, b\) forms an Arithmetic Progression (AP).

The objective is to determine the values of \(a_1, a_2, a_3, \ldots, a_n\) that satisfy this criterion, given the initial and final terms \(a\) and \(b\), respectively, and the number of means \(n\) to be inserted.

The first crucial step in this process involves calculating the common difference (\(d\)) of the AP. This is achieved by utilizing the position of the last term (\(b\)) in the progression. Given that inserting \(n\) means results in \(n+2\) terms (counting both \(a\) and \(b\)), the position of \(b\) is \(n+2\). Therefore, the formula to find \(d\) leverages the fact that \(b\) is the \((n+2)\)-th term in the sequence:

\[b = a + (n+2-1)d\]

Simplifying this formula gives:

\[d = \frac{b - a}{n+1}\]

This calculation ensures that \(d\) is the step size that evenly spaces each term in the progression from \(a\) to \(b\).

With the common difference (\(d\)) determined, the next step is to calculate the value of each arithmetic mean in the sequence. The \(k\)-th arithmetic mean (\(a_k\)), where \(k\) ranges from 1 to \(n\), can be found using the formula:

\[a_k = a + kd\]

This formula efficiently calculates the value of each mean by adding \(k\) times the common difference to the initial term \(a\). It ensures that each \(a_k\) is correctly positioned in the AP, maintaining the constant interval \(d\) between successive terms.

Example

Problem Statement:

Insert 5 arithmetic means (AM's) between the numbers 2 and 100.

Solution:

To insert 5 AM's between the numbers 2 and 100, we need to determine the values of these means such that they, along with the numbers 2 and 100, form an arithmetic progression (AP). Let's denote these means as \(a_1, a_2, a_3, a_4,\) and \(a_5\).

Step 1: Calculate the Common Difference (\(d\))

First, we calculate the common difference (\(d\)) of the AP using the formula:

\[d = \frac{b - a}{n + 1}\]

where:

\(a\) is the first term,
\(b\) is the last term,
\(n\) is the number of arithmetic means to be inserted.

Substituting \(a = 2\), \(b = 100\), and \(n = 5\) into the formula gives:

\[d = \frac{100 - 2}{5 + 1} = \frac{98}{6} = 16.\frac{1}{3}\]

Step 2: Determine Each Arithmetic Mean (\(a_k\))

Next, we find each arithmetic mean using the formula:

[a_k = a + k \cdot d] where \(k\) is the position of the arithmetic mean between \(a\) and \(b\).

For \(a_1\):

\[a_1 = 2 + 1 \cdot 16.\frac{1}{3} = 2 + 16.\frac{1}{3} = 18.\frac{1}{3}\]

For \(a_2\):

\[a_2 = 2 + 2 \cdot 16.\frac{1}{3} = 2 + 32.\frac{2}{3} = 34.\frac{2}{3}\]

For \(a_3\):

\[a_3 = 2 + 3 \cdot 16.\frac{1}{3} = 2 + 49 = 51\]

For \(a_4\):

\[a_4 = 2 + 4 \cdot 16.\frac{1}{3} = 2 + 65.\frac{1}{3} = 67.\frac{1}{3}\]

For \(a_5\):

\[a_5 = 2 + 5 \cdot 16.\frac{1}{3} = 2 + 81.\frac{2}{3} = 83.\frac{2}{3}\]

Conclusion:

The 5 arithmetic means to be inserted between the numbers 2 and 100, to form an arithmetic progression, are \(18.\frac{1}{3}\), \(34.\frac{2}{3}\), \(51\), \(67.\frac{1}{3}\), and \(83.\frac{2}{3}\).

Average of n-AM's

The average of \(n\) arithmetic means (AM's) inserted between two numbers \(a\) and \(b\) is equal to the average of \(a\) and \(b\), which is also equivalent to the single arithmetic mean between \(a\) and \(b\).

Proof:

Given:

Two numbers \(a\) and \(b\), where \(a < b\).
\(n\) arithmetic means \(a_1, a_2, \ldots, a_n\) inserted between \(a\) and \(b\) such that the sequence \(a, a_1, a_2, \ldots, a_n, b\) forms an arithmetic progression (AP).

Let \(d\) be the common difference of the AP.

Calculation of the Common Difference (\(d\))

The common difference \(d\) is given by:

\[d = \frac{b - a}{n + 1}\]

Expression for Each Arithmetic Mean (\(a_k\))

Each arithmetic mean \(a_k\) can be expressed as:

\[a_k = a + k \cdot d\]

where \(k = 1, 2, \ldots, n\).

Average of the \(n\) AM's

The average of the \(n\) AM's is calculated as:

\[\frac{a_1 + a_2 + \ldots + a_n}{n}\]

Substituting the expression for each \(a_k\) into the average formula gives:

\[\frac{1}{n} \left( \sum_{k=1}^{n} (a + k \cdot d) \right)\]

Expanding the summation:

\[\frac{1}{n} \left( n \cdot a + d \cdot \sum_{k=1}^{n} k \right)\]

The sum of the first \(n\) natural numbers is \(\frac{n(n + 1)}{2}\), so:

\[\frac{1}{n} \left( n \cdot a + d \cdot \frac{n(n + 1)}{2} \right)\]

Substituting \(d = \frac{b - a}{n + 1}\) into the equation:

\[\frac{1}{n} \left( n \cdot a + \frac{b - a}{n + 1} \cdot \frac{n(n + 1)}{2} \right)\]

Simplifying:

\[\frac{1}{n} \left( n \cdot a + \frac{n(b - a)}{2} \right)\]

\[\frac{1}{n} \left( \frac{2na + n(b - a)}{2} \right)\]

\[\frac{1}{n} \left( \frac{n(a + b)}{2} \right)\]

\[\frac{a + b}{2}\]

Conclusion

The average of \(n\) arithmetic means inserted between two numbers \(a\) and \(b\) is indeed equal to the average of \(a\) and \(b\), which is the same as the single arithmetic mean between \(a\) and \(b\).

Inserting \(n\) Geometric Means Between Two Positive Numbers \(a\) and \(b\)

When we talk about inserting \(n\) geometric means between two positive numbers \(a\) and \(b\), we are defining a process to find \(n\) numbers \(g_1, g_2, ..., g_n\) such that the sequence \(a, g_1, g_2, ..., g_n, b\) forms a geometric progression (GP). In a GP, each term after the first is obtained by multiplying the preceding term by a constant called the common ratio (\(r\)). Our goal is to determine these \(n\) geometric means so that they fit seamlessly into the GP that starts with \(a\) and ends with \(b\).

Finding the Common Ratio (\(r\))

To determine the \(n\) geometric means, the first step is to calculate the common ratio (\(r\)) of the GP. This is achieved by observing the total number of terms in the sequence, which includes the original numbers \(a\) and \(b\), plus the \(n\) means, making \(n+2\) terms in total.

Given \(a\) as the first term and \(b\) as the last term of the GP, and recognizing that there are \(n+1\) intervals (or steps) from \(a\) to \(b\), the formula to find \(b\) in terms of \(a\), \(r\), and \(n\) is:

\[b = a \cdot r^{(n+2-1)}\]

\[b = a \cdot r^{(n+1)}\]

This equation reflects that starting from \(a\), we multiply by the common ratio \(r\) a total of \(n+1\) times to reach \(b\).

Rearranging this formula to solve for \(r\) yields:

\[r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}\]

This calculation for \(r\) reveals how to scale \(a\) incrementally across \(n+1\) steps to ultimately reach \(b\), ensuring each step is uniform in terms of the GP.

Calculating Each Geometric Mean (\(g_k\))

With the common ratio \(r\) now known, we can calculate the value of each geometric mean in the sequence. Since, \(g_1\) is the second term, \(g_1 = ar\), then \(g_2=ar^2\) and so on. In general,

\[g_k = a \cdot r^k\]

Product of n-GM's

Given:

Two positive numbers \(a\) and \(b\), with \(a < b\).
\(n\) geometric means \(g_1, g_2, \ldots, g_n\) are to be inserted between \(a\) and \(b\) such that the sequence \(a, g_1, g_2, \ldots, g_n, b\) forms a geometric progression (GP).

Let \(r\) be the common ratio of the GP.

Calculation of the Common Ratio (\(r\))

The common ratio \(r\) is determined by the relationship:

\[r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}\]

Expression for Each Geometric Mean (\(g_k\))

Each geometric mean \(g_k\) in the sequence can be expressed as:

\[g_k = a \cdot r^k\]

where \(k = 1, 2, \ldots, n\).

Product of the \(n\) GM's

The product of the \(n\) GM's can be calculated as:

\[g_1 \cdot g_2 \cdot \ldots \cdot g_n\]

Substituting the expression for each \(g_k\) gives:

\[= a^n \cdot r^{\frac{n(n+1)}{2}}\]

Given \(r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}\), we substitute \(r\) into the product formula:

\[= a^n \cdot \left(\left(\frac{b}{a}\right)^{\frac{1}{n+1}}\right)^{\frac{n(n+1)}{2}}\]

\[= a^n \cdot \left(\frac{b}{a}\right)^{\frac{n}{2}}\]

\[= a^n \cdot b^{\frac{n}{2}} \cdot a^{-\frac{n}{2}}\]

\[= a^{\frac{n}{2}} \cdot b^{\frac{n}{2}}\]

\[= (a \cdot b)^{\frac{n}{2}}\]

Since \(n=1\) corresponds to inserting a single GM between \(a\) and \(b\), the formula simplifies to the geometric mean of \(a\) and \(b\), which is \(\sqrt{ab}\). For \(n>1\), the product of the \(n\) GM's still reflects the relationship between \(a\) and \(b\), rooted in their geometric connection.

n-HM's between \(a\) and \(b\)

Inserting \(n\) Harmonic Means (\(n\)-HMs) between two positive numbers \(a\) and \(b\) involves finding \(n\) numbers \(h_1, h_2, \ldots, h_n\) such that \(a, h_1, h_2, \ldots, h_n, b\) are in Harmonic Progression (HP). To achieve this, we exploit the property that if these numbers are in HP, then their reciprocals \(1/a, 1/h_1, 1/h_2, \ldots, 1/h_n, 1/b\) are in Arithmetic Progression (AP).

Given that \(1/h_1, 1/h_2, \ldots, 1/h_n\) are \(n\) Arithmetic Means (AMs) between \(1/a\) and \(1/b\), the common difference \(d\) of this AP can be expressed as:

\[d = \frac{1/b - 1/a}{n+1}\]

This formula for \(d\) follows from the general formula for finding the common difference in an AP when \(n\) terms are inserted between two given terms.

Thus, the \(k\)th term in this sequence, corresponding to \(1/h_k\), can be written as:

\[1/h_k = 1/a + kd\]

Substituting the value of \(d\) into this expression gives:

\[1/h_k = 1/a + k\left(\frac{1/b - 1/a}{n+1}\right)\]

This formula allows us to calculate each \(h_k\) by inverting the equation:

\[h_k = \frac{1}{1/a + k\left(\frac{1/b - 1/a}{n+1}\right)}\]

This method provides a systematic way to find the \(n\) harmonic means between any two positive numbers \(a\) and \(b\), utilizing the concept of arithmetic progression in the reciprocals of the HP terms.

Example

Problem: Insert 3 Harmonic Means (HMs) between 2 and 10.

Solution: Given:

Two numbers: \(a = 2\), \(b = 10\)
Number of HMs to insert: \(n = 3\)

First, calculate the common difference \(d\) in the Arithmetic Progression (AP) of the reciprocals:

\[d = \frac{1/b - 1/a}{n+1} = \frac{1/10 - 1/2}{3 + 1}\]

\[d = \frac{1/10 - 5/10}{4} = \frac{-4/10}{4} = \frac{-1}{10}\]

To find each harmonic mean \(h_k\), we use the formula for the \(k\)th term in the AP of reciprocals:

\[1/h_k = 1/a + k\cdot d\]

Given \(d = -1/10\), let's calculate \(h_1\), \(h_2\), and \(h_3\) with the corrected value of \(d\).

For \(h_1\) (\(k = 1\)):

\[1/h_1 = 1/2 + 1(-1/10) = 1/2 - 1/10 = 5/10 - 1/10 = 4/10 = 2/5\]

\[h_1 = 5/2 = 2.5\]

For \(h_2\) (\(k = 2\)):

\[1/h_2 = 1/2 + 2(-1/10) = 1/2 - 2/10 = 5/10 - 2/10 = 3/10\]

\[h_2 = 10/3 \approx 3.33\]

For \(h_3\) (\(k = 3\)):

\[1/h_3 = 1/2 + 3(-1/10) = 1/2 - 3/10 = 5/10 - 3/10 = 2/10 = 1/5\]

\[h_3 = 5\]

Therefore, the harmonic means to be inserted between 2 and 10 are \(h_1 = 2.5\), \(h_2 \approx 3.33\), and \(h_3 = 5\).