Problem 74 - Calculus Early Transcendentals: Pearson New International Edition

Question

For two positive numbers \(a\) and \(b\), prove that $$ \sqrt{a b} \leq \frac{1}{2}(a+b) $$ This is the simplest version of a famous inequality called the geometric mean-arithmetic mean inequality.

Step-by-Step Solution

Answer

\( \sqrt{ab} \leq \frac{1}{2}(a + b) \) is always true for positive \( a \) and \( b \).

Step 1: Understand the Inequality

We are given the inequality \( \sqrt{ab} \leq \frac{1}{2}(a + b) \) and need to prove it. This inequality states that the geometric mean of two positive numbers \( a \) and \( b \) is less than or equal to their arithmetic mean.

Step 2: Represent the Inequality Differently

Let's rearrange the inequality \( \sqrt{ab} \leq \frac{1}{2}(a + b) \) by squaring both sides to eliminate the square root. This is valid since both sides are non-negative for positive \( a \) and \( b \). So, we have: \( ab \leq \frac{1}{4}(a + b)^2 \).

Step 3: Expand and Simplify the Right Side

Now expand the right side: \( \frac{1}{4}(a + b)^2 = \frac{1}{4}(a^2 + 2ab + b^2) = \frac{1}{4}a^2 + \frac{1}{2}ab + \frac{1}{4}b^2 \).

Step 4: Check the Equivalent Inequality

Substitute the expansion into the inequality: \( ab \leq \frac{1}{4}a^2 + \frac{1}{2}ab + \frac{1}{4}b^2 \). After simplifying by subtracting \( ab \) from both sides, we get: \( 0 \leq \frac{1}{4}a^2 - \frac{1}{2}ab + \frac{1}{4}b^2 \).

Step 5: Recognize a Perfect Square

Notice that the expression \( \frac{1}{4}a^2 - \frac{1}{2}ab + \frac{1}{4}b^2 \) can be rewritten as \( \left( \frac{1}{2}a - \frac{1}{2}b \right)^2 \).

Step 6: Conclude with Inequality Validity

Since \( \left( \frac{1}{2}a - \frac{1}{2}b \right)^2 \geq 0 \) for all real numbers \( a \) and \( b \), the given inequality is verified. Hence, \( \sqrt{ab} \leq \frac{1}{2}(a + b) \) is true.

Key Concepts

Geometric Mean

The geometric mean is a concept used to define a central tendency of a set of numbers—specifically, positive numbers. When you have two numbers, say \( a \) and \( b \), the geometric mean is calculated by taking the product of the numbers and then finding the square root of that product. To illustrate, for two numbers \( a \) and \( b \), the geometric mean is given by \( \sqrt{ab} \). It tends to provide a better central value when numbers multiply to a single output, such as compound interest or growth rates.

• The geometric mean is always less than or equal to the arithmetic mean when dealing with positive numbers. This is the root of the AM-GM inequality.
• It's particularly useful when dealing with numbers of different magnitudes, as it reduces the impact of very high numbers compared to very low numbers.

Understanding the geometric mean can help you better appreciate the balance and proportions between related quantities.

Arithmetic Mean

The arithmetic mean, commonly known as the average, is the sum of a set of numbers divided by the count of numbers in the set. It is a straightforward measure of central tendency and provides a quick snapshot of a data set's central value.For two numbers, \( a \) and \( b \), the arithmetic mean is calculated as \( \frac{1}{2}(a + b) \). In contrast to the geometric mean, the arithmetic mean incorporates every individual value uniformly, making it a suitable measure when all numbers should contribute equally.

• The arithmetic mean is very easy to compute and understand, especially in educational and practical scenarios.
• Unlike the geometric mean, the arithmetic mean can be affected significantly by very high or low numbers, sometimes leading to misleading representations of data.

Familiarity with the arithmetic mean is invaluable as it is fundamental to statistics and everyday numerical evaluation.

Inequality Proof

The proof of inequalities, like the famous AM-GM inequality, often involves algebraic manipulation to reveal properties of numbers that are not immediately obvious. In the context of the exercise, we need to prove that the geometric mean of two positive numbers is less than or equal to their arithmetic mean.The process involves:

• Starting from the given inequality \( \sqrt{ab} \leq \frac{1}{2}(a + b) \).
• Squaring both sides, resulting in \( ab \leq \frac{1}{4}(a^2 + 2ab + b^2) \).
• Simplifying the expanded form to demonstrate that the expression \( \frac{1}{4}a^2 - \frac{1}{2}ab + \frac{1}{4}b^2 \) is a perfect square.

Finally, recognizing that \( \left( \frac{1}{2}a - \frac{1}{2}b \right)^2 \geq 0 \) offers essential insight:

• This means that any squared term is always non-negative, confirming the inequality holds for all positive \( a \) and \( b \).

Understanding this proof and others like it sharpens mathematical comprehension and skills in logic and algebraic reasoning.