Problem 60 - Calculus Early Transcendentals

Question

Maximizing a sum Geometric and arithmetic means Prove that the geometric mean of a set of positive numbers \(\left(x_{1} x_{2} \cdots x_{n}\right)^{1 / n}\) is no greater than the arithmetic mean \(\left(x_{1}+\cdots+x_{n}\right) / n\) in the following cases. a. Find the maximum value of \(x y z,\) subject to \(x+y+z=k\) where \(k\) is a real number and \(x>0, y>0,\) and \(z>0 .\) Use the result to prove that $$(x y z)^{1 / 3} \leq \frac{x+y+z}{3}$$ b. Generalize part (a) and show that $$\left(x_{1} x_{2} \cdots x_{n}\right)^{1 / n} \leq \frac{x_{1}+\cdots+x_{n}}{n}$$

Step-by-Step Solution

Answer

Based on the given step-by-step solution, answer the following question: Question: Prove that for any positive integers x, y, and z, the inequality \((xyz)^{1/3} \leq \frac{x+y+z}{3}\) holds. Answer: Based on the AM-GM inequality, we know that the geometric mean of a set of positive numbers is never greater than the arithmetic mean. In the case of three positive numbers x, y, and z, we can apply the inequality as follows: \(\frac{x+y+z}{3} \geq \sqrt[3]{xyz}\) We are also given that x + y + z = k. Therefore, we can rewrite the inequality in terms of k: \(\frac{k}{3} \geq \sqrt[3]{xyz}\) Thus, the inequality is proven as \((xyz)^{1/3} \leq \frac{x+y+z}{3}\).

Step 1: Apply the AM-GM inequality for n=3

According to the AM-GM inequality, the geometric mean of a set of positive numbers is never greater than the arithmetic mean. In the case of three positive numbers x, y, and z, we can apply the AM-GM inequality as follows: \(\frac{x+y+z}{3} \geq \sqrt[3]{xyz}\)

Step 2: Use the constraint x+y+z=k

We are given that the sum x+y+z=k. Therefore, we can rewrite the inequality in terms of k: \(\frac{k}{3} \geq \sqrt[3]{xyz}\)

Step 3: Find the maximum value of xyz

In order to find the maximum value of the product xyz under the given constraint, we will reverse the inequality and cube both sides: \((xyz) \leq \left(\frac{k}{3}\right)^3\) We can see that the maximum value of xyz is \(\left(\frac{k}{3}\right)^3\) when the inequality holds.

Step 4: Prove the desired inequality for n=3

Now, we can show that \((xyz)^{1/3} \leq \frac{x+y+z}{3}\): \((xyz)^{1/3} \leq \sqrt[3]{\left(\frac{k}{3}\right)^3} = \frac{k}{3} = \frac{x+y+z}{3}\) We have successfully proved the inequality for n=3. #Part (b): Generalizing the result for any positive integer n#

Step 5: Apply the AM-GM inequality for n positive numbers

Similar to the case of n=3, we apply the AM-GM inequality for any positive integer n: \(\frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1x_2\cdots x_n}\)

Step 6: Rearrange the inequality

Now, we want to rearrange the inequality to match the desired inequality: \(\left(x_1x_2\cdots x_n\right)^{1/n} \leq \frac{x_1 + x_2 + \cdots + x_n}{n}\) We have successfully generalized the result for any positive integer n. Thus, we have proved that the geometric mean of a set of positive numbers is no greater than the arithmetic mean.

Key Concepts

Geometric Mean

The geometric mean is a measure that captures the central tendency of a set of numbers by using their product. For any set of positive numbers \( x_1, x_2, \ldots, x_n \), the geometric mean is calculated by taking the \( n^{th} \) root of their product:

• Geometric Mean: \( \sqrt[n]{x_1 \cdot x_2 \cdot \cdots \cdot x_n} \)

This type of mean is particularly useful in cases where the numbers are of different scales, such as growth rates or financial returns, because it encapsulates the multiplicative relation between the numbers efficiently. It shows the "average" multiplicative rate for the entire set of numbers.

Arithmetic Mean

The arithmetic mean, often just called the "average," is a widely used measure of central tendency. It is calculated by summing a list of numbers and then dividing the sum by the count of those numbers:

• Arithmetic Mean: \( \frac{x_1 + x_2 + \cdots + x_n}{n} \)

This mean is intuitive and straightforward for understanding the typical value in a set of numbers. It treats all values with equal weight, which makes it ideal in scenarios where equal importance is attached to all the numbers in the dataset.

Inequality Proof

The inequality proof in question demonstrates the relationship between the geometric mean and arithmetic mean using the well-known AM-GM inequality. This inequality states that for any set of non-negative numbers, the arithmetic mean is always equal to or greater than the geometric mean:

• AM-GM Inequality: \( \frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 \cdot x_2 \cdot \cdots \cdot x_n} \)

Why the Inequality Holds

The inequality ensures that when comparing sets of numbers, the arithmetic mean, calculated through simple addition and division, will always encompass or exceed the geometric mean. This encapsulates the tendency of arithmetic mean to account for all values, regardless of their distribution or relative size. Hence, the AM-GM inequality becomes a crucial tool for comparing average values and understanding the structure of data.

Maximization Problem

In the context of the AM-GM inequality, the maximization problem involves determining the highest possible product of a set of positive numbers given a fixed sum of those numbers. In this case, we are given that \( x + y + z = k \), where \( x, y, z > 0 \). We aim to maximize the product \( xyz \):

• Maximum Product: \( xyz \leq \left( \frac{k}{3} \right)^3 \)

Why Equal Distribution Works Best

The strategy to find this maximum relies on equally partitioning "k" for each variable \( x, y, z \) such that all variables are the same: \( x = y = z = \frac{k}{3} \). This alignment results in achieving the maximum product, as equal distribution tends to harmonize the balance between sum and multiplication among the numbers. By applying the AM-GM inequality, we validate that this uniform distribution yields the highest possible product under the given constraints.