Problem 69 - Calculus Early Transcendentals: Pearson New International Edition

Question

The arithmetic mean of the numbers \(a\) and \(b\) is \((a+b) / 2\), and the geometric mean of two positive numbers \(a\) and \(b\) is \(\sqrt{a b} .\) Suppose that \(a>0\) and \(b>0\). (a) Show that \(\sqrt{a b} \leq(a+b) / 2\) holds by squaring both sides and simplifying. (b) Use calculus to show that \(\sqrt{a b} \leq(a+b) / 2 .\) Hint: Consider \(a\) to be fixed. Square both sides of the inequality and divide through by \(b .\) Define the function \(F(b)=(a+b)^{2} / 4 b\). Show that \(F\) has its minimum at \(a\). (c) The geometric mean of three positive numbers \(a, b\), and \(c\) is \((a b c)^{1 / 3} .\) Show that the analogous inequality holds: $$ (a b c)^{1 / 3} \leq \frac{a+b+c}{3} $$ Hint: Consider \(a\) and \(c\) to be fixed and define \(F(b)=\) \((a+b+c)^{3} / 27 b .\) Show that \(F\) has a minimum at \(b=\) \((a+c) / 2\) and that this minimum is \([(a+c) / 2]^{2}\). Then use the result from (b).

Step-by-Step Solution

Answer

The inequality \(\sqrt{ab} \leq \frac{a+b}{2}\) holds for positive \(a\) and \(b\). For three numbers, \((abc)^{1/3} \leq \frac{a+b+c}{3}\).

Step 1: Understand the Inequality

We need to demonstrate that the geometric mean of two numbers, \(\sqrt{ab}\), is less than or equal to the arithmetic mean \((a+b)/2\). This inequality is represented as \(\sqrt{ab} \leq \frac{a+b}{2}\).

Step 2: Square Both Sides (a)

Square both sides of the inequality to eliminate the square root: \( ab \leq \left(\frac{a+b}{2}\right)^2 \). The right-hand side becomes \( \frac{(a+b)^2}{4} = \frac{a^2 + 2ab + b^2}{4} \). Thus, verify \(4ab \leq a^2 + 2ab + b^2\).

Step 3: Simplify the Inequality (a)

Rewriting the inequality gives \(4ab \leq a^2 + 2ab + b^2\). Bring all terms to one side: \(0 \leq a^2 - 2ab + b^2\). This expression is \((a-b)^2\), which is always non-negative.

Step 4: Utilize Calculus (b)

Define a function for a fixed \(a\): \(F(b) = \frac{(a+b)^2}{4b}\). Find its derivative with respect to \(b\).

Step 5: Derivative and Critical Point (b)

Differentiate \(F(b)\): \( F'(b) = \frac{(b-a)^2}{4b^2} \). The critical point occurs when \(b = a\).

Step 6: Verify Minimum at Critical Point (b)

Evaluate \(F(b)\) at \(b = a\) to confirm it is a minimum point. Check the second derivative \(F''(b)\) to confirm it is positive at \(b = a\).

Step 7: Extend to Three Variables (c)

Consider \(a, c\) as constants and define \(F(b) = \frac{(a+b+c)^3}{27b}\). Use the first and second derivatives to find critical points and confirm minimum at \(b = \frac{a+c}{2}\).

Step 8: Conclusion for Three Variables (c)

Evaluate \(F(b)\) at the minimum and apply results to show \((abc)^{1/3} \leq \frac{a+b+c}{3}\) utilizing the earlier inequality \(\sqrt{ab} \leq \frac{a+b}{2}\).

Key Concepts

Calculus Optimization

Calculus Optimization helps us find the maximum or minimum values of a function, which is useful for demonstrating inequalities. In our exercise, we applied calculus optimization to show the Arithmetic and Geometric Means inequality. By defining a function involving a parameter (i.e., \(F(b) = \frac{(a+b)^2}{4b}\) for a fixed value \(a\)), we can use calculus to determine where the minima occur.
First, we take the derivative of the function with respect to \(b\), the variable of interest, to find critical points. For instance, the derivative of \(F(b)\) leads us to the critical point \(b = a\).
By performing a derivative test (such as the second derivative test), we confirm if this point is a minimum. Here, we found \(F'(b) = \frac{(b-a)^2}{4b^2}\), and by evaluating the second derivative \(F''(b)\), we confirmed it is positive for \(b = a\), indicating a minimum.
This mathematical technique proves that in the context of optimization, our inequality \(\sqrt{ab} \leq \frac{a+b}{2}\) holds, making calculus a powerful tool in proving inequalities.

Inequality Proof

Proving inequalities often involves transforming them into different expressions or forms that are easier to handle. An essential part of understanding arithmetic and geometric mean inequalities is recognizing how algebraic manipulations and calculus can be applied.
For example, the problem begins by taking the inequality we want to prove: the geometric mean \(\sqrt{ab}\) is less than or equal to the arithmetic mean \((a+b)/2\). We then square both sides of the inequality, which helps eliminate the square root:

• Square \(\sqrt{ab}\) gives \(ab\).
• Square \((a+b)/2\) results in \(\frac{a^2 + 2ab + b^2}{4}\).

Rewriting as \(4ab \leq a^2 + 2ab + b^2\) helps simplify further by rearranging the terms: \(0 \leq (a-b)^2\), which is always true, since a square of a real number is non-negative.
The method shows that mathematical inequalities often rely on reformulating expressions so it's easier to demonstrate their truth.

Geometric Mean

The Geometric Mean is a type of average which is especially useful for multiplying different factors together, often appearing in growth rates and proportions.
For two numbers \(a\) and \(b\), it is defined as \(\sqrt{ab}\). This mean emphasizes the "multiplicative" relationship between numbers, offering a measure of central tendency meaningful even if the numbers vary significantly.
In the context of inequalities, the geometric mean exhibits an interesting relationship with the arithmetic mean; it is always less than or equal to the arithmetic mean, as demonstrated through our exercise of mathematical proof and optimization. For three numbers \(a, b,\) and \(c\), it's expressed as \((abc)^{1/3}\).
The process shows the versatility and effectiveness of the geometric mean when used in proving inequalities like \(\sqrt{ab} \leq (a+b)/2\). It emphasizes numbers' relationships in a multiplicative sense, providing insight into their proportional arrangements.

Arithmetic Mean

The Arithmetic Mean, commonly known as the average, is simply the sum of numbers divided by the count of numbers. It represents a "central" value for a data set.
For two numbers \(a\) and \(b\), it is expressed as \((a+b)/2\). This mean provides a straightforward measure of center that balances the numbers, offering a simple perspective on their collective value.
In proving the inequality between arithmetic and geometric means, the arithmetic mean's property of summing values creates an upper bound for comparisons. It is globally applicable due to this balancing characteristic.
This property comes elegantly into play when applying it to sets of two or three values; the arithmetic mean offers an upper benchmark that the geometric mean cannot exceed, as was shown in our exercise. This showcases why the arithmetic mean is essential in statistical calculations and comparisons.