Proof of the First Inequality

Oct 13, 2025 by sorusgentrification.com 39 views

Introduction

In the realm of real analysis, sequences and series are fundamental concepts that have numerous applications in mathematics and other fields. Given two sequences of $n$ positive real numbers: $\left( a_{i} \right)_{i=1}^n$ and $\left( b_{i} \right)_{i=1}^n$ , such that they have the same sum, i.e. $$\sum_{i=1}^{{n}a_{i}=\sum_{i=1}}{n}b_{i}$$ we are interested in comparing the product of these sequences when translated by a fixed positive number. This problem has significant implications in various areas of mathematics, including algebra precalculus and inequality.

Problem Formulation

Let $\left( a_{i} \right)_{i=1}^n$ and $\left( b_{i} \right)_{i=1}^n$ be two sequences of $n$ positive real numbers, and let $c$ be a fixed positive number. We are interested in comparing the product of these sequences when translated by $c$ , i.e., we want to compare the values of $\prod_{i=1}^{n} (a_i + c)$ and $\prod_{i=1}^{n} (b_i + c)$ .

Theoretical Background

To approach this problem, we need to recall some fundamental concepts from real analysis, including the properties of sequences and series. Specifically, we will use the following results:

The sum of a sequence is equal to the sum of its terms.
The product of a sequence is equal to the product of its terms.
The inequality of arithmetic and geometric means (AM-GM inequality).

Comparison of Products

We will now compare the products of the sequences $\left( a_{i} \right)_{i=1}^n$ and $\left( b_{i} \right)_{i=1}^n$ when translated by $c$ . We will use the following approach:

We will first show that the product of the sequence $\left( a_{i} \right)_{i=1}^n$ when translated by $c$ is greater than or equal to the product of the sequence $\left( b_{i} \right)_{i=1}^n$ when translated by $c$ .
We will then show that the product of the sequence $\left( b_{i} \right)_{i=1}^n$ when translated by $c$ is greater than or equal to the product of the sequence $\left( a_{i} \right)_{i=1}^n$ when translated by $c$ .

Proof of the First Inequality

We will now prove the first inequality, i.e., we will show that the product of the sequence $\left( a_{i} \right)_{i=1}^n$ when translated by $c$ is greater than or equal to the product of the sequence $\left( b_{i} \right)_{i=1}^n$ when translated by $c$ .

By the AM-GM inequality, we have:

\frac{a_i + c}{b_i + c} \geq 1

for all $i = 1, 2, \ldots, n$ . Multiplying both sides of this inequality by itself $n$ times, we get:

\prod_{i=1}^{n} \frac{a_i + c}{b_i + c} \geq 1

Rearranging this inequality, we get:

\prod_{i=1}^{n} (a_i + c) \geq \prod_{i=1}^{n} (b_i + c)

This proves the first inequality.

Proof of the Second Inequality

We will now prove the second inequality, i.e., we will show that the product of the sequence $\left( b_{i} \right)_{i=1}^n$ when translated by $c$ is greater than or equal to the product of the sequence $\left( a_{i} \right)_{i=1}^n$ when translated by $c$ .

By the AM-GM inequality, we have:

\frac{b_i + c}{a_i + c} \geq 1

for all $i = 1, 2, \ldots, n$ . Multiplying both sides of this inequality by itself $n$ times, we get:

\prod_{i=1}^{n} \frac{b_i + c}{a_i + c} \geq 1

Rearranging this inequality, we get:

\prod_{i=1}^{n} (b_i + c) \geq \prod_{i=1}^{n} (a_i + c)

This proves the second inequality.

Conclusion

In this article, we have compared the product of two sequences of $n$ positive real numbers when translated by a fixed positive number. We have shown that the product of the sequence $\left( a_{i} \right)_{i=1}^n$ when translated by $c$ is greater than or equal to the product of the sequence $\left( b_{i} \right)_{i=1}^n$ when translated by $c$ , and vice versa. This result has significant implications in various areas of mathematics, including algebra precalculus and inequality.

References

[1] Hardy, G. H., & Littlewood, J. E. (1914). Some properties of trigonometric series. Messenger of Mathematics, 43, 134-146.
[2] Mitrinović, D. S. (1968). Analytic Inequalities. Springer-Verlag.
[3] Beckenbach, E. F., & Bellman, R. (1961). Inequalities. Springer-Verlag.

Future Work

In future work, we plan to extend this result to more general sequences and to investigate the implications of this result in various areas of mathematics. We also plan to explore the use of this result in applications such as optimization and machine learning.

Appendix

In this appendix, we provide some additional results and examples that are related to the main result of this article.

Example 1

Let $\left( a_{i} \right)_{i=1}^n$ and $\left( b_{i} \right)_{i=1}^n$ be two sequences of $n$ positive real numbers, and let $c$ be a fixed positive number. Suppose that $\sum_{i=1}^{n}a_{i}=\sum_{i=1}^{n}b_{i}$ and $\prod_{i=1}^{n} (a_i + c) = \prod_{i=1}^{n} (b_i + c)$ . Show that $a_i = b_i$ for all $i = 1, 2, \ldots, n$ .

Solution

We have:

\prod_{i=1}^{n} (a_i + c) = \prod_{i=1}^{n} (b_i + c)

Taking the logarithm of both sides of this equation, we get:

\sum_{i=1}^{n} \log (a_i + c) = \sum_{i=1}^{n} \log (b_i + c)

Using the property of logarithms that $\log x - \log y = \log \frac{x}{y}$ , we get:

\sum_{i=1}^{n} \log \frac{a_i + c}{b_i + c} = 0

Since the logarithm function is continuous and differentiable, we can differentiate both sides of this equation with respect to $c$ to get:

\sum_{i=1}^{n} \frac{1}{a_i + c} = \sum_{i=1}^{n} \frac{1}{b_i + c}

Since $c$ is a fixed positive number, we can choose $c$ such that $a_i + c \neq 0$ and $b_i + c \neq 0$ for all $i = 1, 2, \ldots, n$ . Then, we can cancel the denominators on both sides of this equation to get:

\sum_{i=1}^{n} 1 = \sum_{i=1}^{n} 1

This implies that $a_i = b_i$ for all $i = 1, 2, \ldots, n$ .

Example 2

Introduction

In our previous article, we compared the product of two sequences of $n$ positive real numbers when translated by a fixed positive number. We showed that the product of the sequence $\left( a_{i} \right)_{i=1}^n$ when translated by $c$ is greater than or equal to the product of the sequence $\left( b_{i} \right)_{i=1}^n$ when translated by $c$ , and vice versa. In this article, we will answer some frequently asked questions related to this result.

Q: What is the significance of this result?

A: This result has significant implications in various areas of mathematics, including algebra precalculus and inequality. It provides a new perspective on the comparison of products of sequences and has potential applications in optimization and machine learning.

Q: Can you provide an example of how this result can be used in real-world applications?

A: Yes, this result can be used in various real-world applications, such as:

Optimization: This result can be used to optimize the product of a sequence of numbers, which is a common problem in optimization.
Machine Learning: This result can be used to improve the performance of machine learning algorithms, such as neural networks, which rely on the product of a sequence of numbers.
Finance: This result can be used to compare the products of two sequences of financial data, such as stock prices or interest rates.

Q: What are the limitations of this result?

A: This result has some limitations, including:

Assumptions: This result assumes that the sequences are composed of positive real numbers, which may not always be the case in real-world applications.
Translation: This result assumes that the sequences are translated by a fixed positive number, which may not always be the case in real-world applications.
Comparison: This result only compares the products of two sequences, which may not always be the case in real-world applications.

Q: Can you provide a proof of this result?

A: Yes, we can provide a proof of this result. The proof is based on the AM-GM inequality and is as follows:

We start by assuming that the sequences are composed of positive real numbers and that the sequences are translated by a fixed positive number.
We then use the AM-GM inequality to show that the product of the sequence $\left( a_{i} \right)_{i=1}^n$ when translated by $c$ is greater than or equal to the product of the sequence $\left( b_{i} \right)_{i=1}^n$ when translated by $c$ , and vice versa.
We then use this result to show that the product of the sequence $\left( a_{i} \right)_{i=1}^n$ when translated by $c$ is greater than or equal to the product of the sequence $\left( b_{i} \right)_{i=1}^n$ when translated by $c$ , and vice versa.

Q: Can you provide a counterexample to this result?

A: Yes, we can provide a counterexample to this result. The counterexample is as follows:

Let $\left( a_{i} \right)_{i=1}^n$ and $\left( b_{i} \right)_{i=1}^n$ be two sequences of $n$ positive real numbers, and let $c$ be a fixed positive number.
Suppose that $\sum_{i=1}^{n}a_{i}=\sum_{i=1}^{n}b_{i}$ and $\prod_{i=1}^{n} (a_i + c) < \prod_{i=1}^{n} (b_i + c)$ .
Then, we can show that there exists an $i$ such that $a_i < b_i$ .

Conclusion

In this article, we have answered some frequently asked questions related to the comparison of product of sequences translated by a fixed positive number. We have shown that this result has significant implications in various areas of mathematics, including algebra precalculus and inequality, and has potential applications in optimization and machine learning. We have also provided a proof of this result and a counterexample to this result.

References

[1] Hardy, G. H., & Littlewood, J. E. (1914). Some properties of trigonometric series. Messenger of Mathematics, 43, 134-146.
[2] Mitrinović, D. S. (1968). Analytic Inequalities. Springer-Verlag.
[3] Beckenbach, E. F., & Bellman, R. (1961). Inequalities. Springer-Verlag.

Future Work

Appendix

In this appendix, we provide some additional results and examples that are related to the main result of this article.

Example 1

Solution

We have:

\prod_{i=1}^{n} (a_i + c) = \prod_{i=1}^{n} (b_i + c)

Taking the logarithm of both sides of this equation, we get:

\sum_{i=1}^{n} \log (a_i + c) = \sum_{i=1}^{n} \log (b_i + c)

Using the property of logarithms that $\log x - \log y = \log \frac{x}{y}$ , we get:

\sum_{i=1}^{n} \log \frac{a_i + c}{b_i + c} = 0

Since the logarithm function is continuous and differentiable, we can differentiate both sides of this equation with respect to $c$ to get:

\sum_{i=1}^{n} \frac{1}{a_i + c} = \sum_{i=1}^{n} \frac{1}{b_i + c}

\sum_{i=1}^{n} 1 = \sum_{i=1}^{n} 1

This implies that $a_i = b_i$ for all $i = 1, 2, \ldots, n$ .

Example 2

Let $\left( a_{i} \right)_{i=1}^n$ and $\left( b_{i} \right)_{i=1}^n$ be two sequences of $n$ positive real numbers, and let $c$ be a fixed positive number. Suppose that $\sum_{i=1}^{n}a_{i}=\sum_{i=1}^{n}b_{i}$ and $\prod_{i=1}^{n} (a_i + c) < \prod_{i=1}^{n} (b_i + c)$ . Show that there exists an $i$ such that $a_i < b_i$ .

Solution

We have:

\prod_{i=1}^{n} (a_i + c) < \prod_{i=1}^{n} (b_i + c)

Taking the logarithm of both sides of this equation, we get:

\sum_{i=1}^{n} \log (a_i + c) < \sum_{i=1}^{n} \log (b_i + c)

Using the property of logarithms that $\log x - \log y = \log \frac{x}{y}$ , we get:

\sum_{i=1}^{n} \log \frac{a_i + c}{b_i + c} < 0

Since the logarithm function is continuous and differentiable, we can differentiate both sides of this equation with respect to $c$ to get:

\sum_{i=1}^{n} \frac{1}{a_i + c} < \sum_{i=1}^{n} \frac{1}{b_i + c}

Since $c$ is a fixed positive number, we can choose $c$ such