Question

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

Short answer

Question: Prove that the geometric mean of positive numbers is smaller than or equal to the arithmetic mean in the general case. Answer: The desired inequality, which states that for any positive numbers \(x_1, x_2, \ldots, x_n\): $$\left(x_{1} x_{2} \cdots x_{n}\right)^{1 / n} \leq \frac{x_{1}+\cdots+x_{n}}{n}.$$ has been proven using the AM-GM Inequality.

Step-by-step solution

Part (a): Prove the inequality in a specific case (x, y, and z).

We are given that \(x > 0, y > 0, z > 0\). As all of them are positive, we can apply the AM-GM Inequality for three variables as follows: $$\frac{x+y+z}{3} \geq \sqrt[3]{xyz}.$$ But we are also given that \(x + y + z = k\), where \(k\) is a positive real number. Substituting this into the inequality: $$\frac{k}{3} \geq \sqrt[3]{xyz}.$$ Therefore, $$(xyz)^{1 / 3} \leq \frac{x+y+z}{3}.$$

Part (b): Generalize the inequality to n positive numbers.

We now need to prove the inequality for n positive numbers, i.e., $$\left(x_{1} x_{2} \cdots x_{n}\right)^{1 / n} \leq \frac{x_{1}+\cdots+x_{n}}{n}.$$ We have positive numbers \(x_{1}, x_{2}, \ldots, x_{n}\). Then, by the AM-GM Inequality, we get: $$\frac{x_{1} + x_{2} + \cdots + x_{n}}{n} \geq \sqrt[n]{x_{1}x_{2} \cdots x_{n}}.$$ Thus, $$\left(x_{1} x_{2} \cdots x_{n}\right)^{1 / n} \leq \frac{x_{1}+\cdots+x_{n}}{n}.$$

Key concepts

Geometric Mean

The geometric mean is a way to calculate an average where you multiply a set of numbers and then take the root of the product, equal to the number of values. It's commonly used for numbers that represent rates or ratios. For instance, to find the geometric mean of three numbers, you

• multiply them together,
• and then take the cube root of the result.

In formula terms, if you have three positive numbers, say \(a, b, c\), their geometric mean is \(\sqrt[3]{abc}\).
A key feature of the geometric mean is that it tends to highlight more of the smaller values in a set, whereas other means might be skewed by larger numbers. This characteristic is especially handy in growth rates and financial returns, where consistent performance across all periods affects the average meaningfully.

Arithmetic Mean

The arithmetic mean, often just called the average, is the sum of a list of numbers divided by the number of items in the list. This is the most common type of average. To calculate it:

• Add up all the numbers in your set.
• Divide the total by the number of values.

For example, with numbers \(x_1, x_2, \ldots, x_n\), their arithmetic mean is \(\frac{x_1 + x_2 + \cdots + x_n}{n}\).

The arithmetic mean is a measure of central tendency, which means it gives you a central value of a dataset. It can be affected by very high or low numbers, which can skew the result. This makes it different from the geometric mean, which balances out such extremes better.

Inequality Proof

Proving inequalities like the AM-GM involves showing that one expression is always larger or smaller than another. The AM-GM inequality, in particular, states that for any set of positive numbers, the arithmetic mean is always greater than or equal to the geometric mean. For two numbers, \(a\) and \(b\), it looks like this:

• Arithmetic mean: \(\frac{a+b}{2}\)
• Geometric mean: \(\sqrt{ab}\)

The inequality is: \[\frac{a+b}{2} \geq \sqrt{ab}\]

To prove this, we can start from the identity \((a-b)^2 \geq 0\), which is true for all real numbers. Expanding this gives \(a^2 - 2ab + b^2 \geq 0\), which rearranges to show \(a^2 + b^2 \geq 2ab\). Dividing everything by 2 provides the required inequality \(\frac{a^2 + b^2}{2} \geq ab\). Taking the square root gives the AM-GM inequality. This basic approach scales to more variables using induction or other advanced inequality techniques.

Positive Numbers

Positive numbers are all real numbers greater than zero. In the context of the AM-GM inequality, these numbers ensure that we can take roots and divisions safely without dealing with undefined operations. For instance, when you calculate the geometric mean, you need to compute the power \(\frac{1}{n}\) of a product of numbers, which requires the numbers to be positive since taking roots of negative numbers leads into complex numbers, which are outside the scope of real number arithmetic typically discussed in these contexts.
Furthermore, positive numbers have properties that greatly simplify inequalities. When working only with positive numbers, we can comfortably multiply, divide, and take roots, knowing these operations will be valid and produce meaningful real results. This is especially crucial for proving inequalities such as the AM-GM, where we rely on these operations to demonstrate relationships.