Question

Let \(w=x_{1} x_{2} \cdots x_{n}\). (a) Maximize \(w\) subject to \(x_{1}+x_{2}+\cdots+x_{n}=1\) and all \(x_{i}>0\) (b) Use part (a) to deduce the famous Geometric Mean\(\begin{array}{llll}\text { Arithmetic } & \text { Mean Inequality for positive numbers }\end{array}\) \(a_{1}, a_{2}, \ldots, a_{n} ;\) that is, $$ \sqrt[n]{a_{1} a_{2} \cdots a_{n}} \leq \frac{a_{1}+a_{2}+\cdots+a_{n}}{n} $$

Short answer

Max product when all terms equal each \( \frac{1}{n} \); deduce AM-GM.

Step-by-step solution

Define the Problem

We need to maximize the product \( w = x_1 x_2 \cdots x_n \) given the constraint \( x_1 + x_2 + \cdots + x_n = 1 \) and \( x_i > 0 \).

Use Lagrange Multipliers

To solve this optimization problem, we use Lagrange multipliers. We set up the function \( f(x_1, x_2, \ldots, x_n, \lambda) = x_1 x_2 \cdots x_n + \lambda(1 - x_1 - x_2 - \cdots - x_n) \).

Compute Partial Derivatives

Calculate the partial derivatives with respect to each \( x_i \) and \( \lambda \). Set them equal to zero:\[\frac{\partial f}{\partial x_i} = \frac{w}{x_i} - \lambda = 0\quad \text{for each } i\text{ from } 1 \text{ to } n,\]\[ \frac{\partial f}{\partial \lambda} = 1 - x_1 - x_2 - \cdots - x_n = 0.\]

Solve the System of Equations

From \( \frac{w}{x_i} = \lambda \), we deduce all \( x_i \) are equal, i.e., \( x_1 = x_2 = \cdots = x_n = x \). Use \( n \times x = 1 \) to get \( x = \frac{1}{n} \). Substitute back to get maximum product: \( w = \left(\frac{1}{n}\right)^n = \frac{1}{n^n} \cdot 1 = \frac{1}{n^n} \).

Prove the AM-GM Inequality

Given positive numbers \( a_1, a_2, \ldots, a_n \), apply the result from part (a) on \( \frac{a_1}{s}, \frac{a_2}{s}, \ldots, \frac{a_n}{s} \) where \( s = a_1 + a_2 + \cdots + a_n \). The product maximization yields\[\prod_{i=1}^n \frac{a_i}{s} \leq \left(\frac{1}{n}\right)^n,\]resulting in\[\prod_{i=1}^n a_i \leq \left( \frac{s}{n} \right)^n,\]which simplifies to the AM-GM inequality:\[\sqrt[n]{a_1 a_2 \cdots a_n} \leq \frac{a_1 + a_2 + \cdots + a_n}{n}.\]

Conclusion

Thus, both parts of the problem are solved by using Lagrange multipliers for maximization and applying it to derive the AM-GM inequality.

Key concepts

Lagrange Multipliers

To solve optimization problems where we have constraints, Lagrange multipliers come in handy. Imagine you want to maximize or minimize a function, but you're restricted by an equation. This is exactly what Lagrange multipliers help with.

Here's the basic idea: You have a function, say, a surface or a similar concept. You also have a constraint, like running along a path. What you do is blend these two things into a new function using a special symbol called lambda (\( \lambda \)). This process involves creating something called a Lagrangian, where you combine your original function with your constraint.

In our problem, the function is the product \( w = x_1 x_2 \ldots x_n \) that we want to maximize. The constraint is the sum of all \( x_i \) being equal to 1. With Lagrange multipliers, you work with the Lagrangian:

• Set up the Lagrangian: \( f(x_1, x_2, \ldots, x_n, \lambda) = x_1 x_2 \cdots x_n + \lambda(1 - x_1 - x_2 - \cdots - x_n) \)
• Take partial derivatives with respect to each variable and \( \lambda \)
• Set these derivatives to zero to find your solution

This way, Lagrange multipliers make tough optimization problems more manageable.

Geometric Mean

Think of the geometric mean (GM) as a way to find the central tendency of a set of numbers. Unlike the usual average (the arithmetic mean), GM multiplies the numbers together and then takes the n-th root, where n is the number total of numbers in the set.

It's particularly useful in situations where we deal with percentages or rates of change. For example, if you're looking at growth rates, the geometric mean gives a more accurate picture than the arithmetic mean.

Let's look at how we calculate the geometric mean for positive numbers \( a_1, a_2, \ldots, a_n \):

• Multiply the numbers: \( a_1 \times a_2 \times \cdots \times a_n \)
• Take the n-th root: \( \sqrt[n]{a_1 a_2 \cdots a_n} \)

So, whenever you encounter a scenario that involves ratios or rates, consider using the geometric mean for a clearer, balanced insight into the data.

Arithmetic Mean

The arithmetic mean (AM), commonly known as the average, is simply the sum of numbers divided by the count of those numbers. It's probably the most familiar way to find the central value of a data set.

Here's how you can calculate the arithmetic mean for a set of values, \( a_1, a_2, \ldots, a_n \):

• Add all the numbers: \( a_1 + a_2 + \cdots + a_n \)
• Divide by the number of values: \( \frac{a_1 + a_2 + \cdots + a_n}{n} \)

The arithmetic mean is straightforward and excels in scenarios where all observations weigh the same. However, it might not always be ideal in skewed distributions or datasets involving varying rates.

AM-GM Inequality

The AM-GM Inequality is a beautiful result in mathematics that shows a relationship between the arithmetic mean and the geometric mean. It says that the arithmetic mean is always greater than or equal to the geometric mean for any set of non-negative numbers.

This is how it can be written for a set of positive numbers \( a_1, a_2, \ldots, a_n \):

• Arithmetic Mean: \( \frac{a_1 + a_2 + \cdots + a_n}{n} \)
• Geometric Mean: \( \sqrt[n]{a_1 a_2 \cdots a_n} \)
• Inequality: \( \sqrt[n]{a_1 a_2 \cdots a_n} \leq \frac{a_1 + a_2 + \cdots + a_n}{n} \)

The inequality is not just a mathematical curiosity. It has practical applications, such as proving economic theories, making investments, and optimizing calculations, among others. It speaks to how averaging with addition can yield different results from averaging with multiplication and roots.

The AM-GM inequality beautifully bridges the arithmetic mean and geometric mean by providing insights into their relationship and capturing the essence of balance in mathematical distributions.