Question

Use the Cauchy inequality to show that \(\sqrt{x y} \leq \frac{1}{2}(x+y)\) for all \(x \geq 0\) and \(y \geq 0\). Here \(\sqrt{x y}\) and \(\frac{1}{2}(x+y)\) are called, respectively, the geometric mean and arithmetic mean of \(x\) and \(y .\) [Hint: Use \(\mathbf{x}=\left[\begin{array}{l}\sqrt{x} \\\ \sqrt{y}\end{array}\right]\) and \(\left.\mathbf{y}=\left[\begin{array}{c}\sqrt{y} \\\ \sqrt{x}\end{array}\right] .\right]\)

Short answer

The inequality is proved using Cauchy-Schwarz and simplifying to \((x-y)^2 \geq 0\).

Step-by-step solution

Set Up the Cauchy-Schwarz Inequality

To use Cauchy-Schwarz, recall the inequality: \((u \cdot v)^2 \leq (u \cdot u)(v \cdot v)\), where \(u\) and \(v\) are vectors. Given \(\mathbf{x}=[\sqrt{x}, \sqrt{y}]^T\) and \(\mathbf{y}=[\sqrt{y}, \sqrt{x}]^T\), we will find the values of \(u \cdot v\), \(u \cdot u\), and \(v \cdot v\).

Calculate the Dot Products

First, calculate \(u \cdot v\) which is \([\sqrt{x}, \sqrt{y}] \cdot [\sqrt{y}, \sqrt{x}] = \sqrt{x} \cdot \sqrt{y} + \sqrt{y} \cdot \sqrt{x} = 2\sqrt{xy}\). Then, calculate \(u \cdot u = [\sqrt{x}, \sqrt{y}] \cdot [\sqrt{x}, \sqrt{y}] = x + y\) and \(v \cdot v = [\sqrt{y}, \sqrt{x}] \cdot [\sqrt{y}, \sqrt{x}] = y + x = x + y\).

Apply the Cauchy-Schwarz Inequality

Substitute the obtained values into the Cauchy-Schwarz inequality: \((2\sqrt{xy})^2 \leq (x+y)(x+y) \). This simplifies to \(4xy \leq (x+y)^2\).

Simplify the Inequality

Expand the right side: \((x+y)^2 = x^2 + 2xy + y^2\). Hence, we have \(4xy \leq x^2 + 2xy + y^2\). Subtract \(4xy\) from both sides to get \(0 \leq x^2 - 2xy + y^2\).

Complete the Square

Recognize that \(x^2 - 2xy + y^2\) can be rewritten as \((x-y)^2\), which is always non-negative. Thus, \(0 \leq (x-y)^2\) is true, concluding that \(\sqrt{xy} \leq \frac{1}{2}(x+y)\).

Key concepts

Geometric Mean

The geometric mean is a measure of the central tendency of two or more numbers. It is different from the arithmetic mean, as it takes the product of the numbers followed by taking the n-th root (where n is the number of values). For two numbers, \(x\) and \(y\), the geometric mean is defined as \(\sqrt{xy}\). This measure is particularly useful in situations where the numbers have a relationship of exponential growth or rates of change, such as in finance or for finding the average rate of return over time.

Unlike the arithmetic mean, which sums up the values before dividing, the geometric mean creates a balanced point by using multiplication, making it suitable for comparing different sets of proportional types of data. In our problem, we compare \(\sqrt{xy}\) with the arithmetic mean to show the relationship between the two for non-negative values of \(x\) and \(y\).

Arithmetic Mean

The arithmetic mean is perhaps the most familiar type of average to most people. It is calculated by adding up a set of numbers and then dividing by the count of those numbers. For the numbers \(x\) and \(y\), the arithmetic mean is given by \(\frac{1}{2}(x+y)\).

This measure is straightforward and shows the "central" number of a grouped set. It is also known as the average. The arithmetic mean assumes that each number in the set contributes equally to the final mean, providing a simple measure of the weight of each number in the context of the set. In the context of the given problem, the arithmetic mean serves as an upper bound for the geometric mean, helping illustrate the principle that the geometric mean can never exceed the arithmetic mean for non-negative real numbers.

Dot Product

The dot product is an operation that takes two equal-length sequences of numbers (usually coordinate vectors), and returns a single number. Algebraically, it is the sum of the products of the corresponding entries of the two sequences.

Given vectors \(\mathbf{u} = [a, b]^T\) and \(\mathbf{v} = [c, d]^T\), the dot product \(\mathbf{u} \cdot \mathbf{v}\) is calculated as \(a \cdot c + b \cdot d\). This concept is crucial in the application of the Cauchy-Schwarz inequality. In our solution, the vectors \(\mathbf{x} = [\sqrt{x}, \sqrt{y}]^T\) and \(\mathbf{y} = [\sqrt{y}, \sqrt{x}]^T\) are used to demonstrate the inequality: \((u \cdot v)^2 \leq (u \cdot u)(v \cdot v)\).

The dot product is valuable as it provides a way to calculate the similarity or the degree of parallelism between two vectors. In geometry, it is often used to compute angles between vectors and lengths of vectors as part of the concept of vector projections.