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 inequalities \( \sqrt{ab} \leq \frac{a+b}{2} \) and \( (abc)^{1/3} \leq \frac{a+b+c}{3} \) are both proven using algebraic manipulation and calculus-derived minima.

Step 1: Start with the expression square both sides

The given inequality is \( \sqrt{a b} \leq \frac{a + b}{2} \). To remove the square root, square both sides to obtain: \( a b \leq \left(\frac{a + b}{2}\right)^2 \).

Step 2: Simplify the inequality

Expand the squared expression on the right-hand side: \[ a b \leq \frac{(a + b)^2}{4} = \frac{a^2 + 2ab + b^2}{4} \]. Multiply both sides by 4 to eliminate the fraction: \( 4ab \leq a^2 + 2ab + b^2 \).

Step 3: Rearrange and factorize

Rearrange the inequality: \( 4ab - 2ab \leq a^2 + b^2 \). Simplify to obtain: \( 2ab \leq a^2 + b^2 \). Reorganize it to \( a^2 - 2ab + b^2 \geq 0 \).

Step 4: Recognize as a perfect square

Notice that \( a^2 - 2ab + b^2 \) can be rewritten as \((a-b)^2 \geq 0\), which is always true since a square of a real number is non-negative.

Step 5: Set up the function F(b) using calculus

Define the function \( F(b) = \frac{(a+b)^2}{4b} \). The goal is to show that this function \( F(b) \) has a minimum at \( b = a \).

Step 6: Differentiate to find critical points

Differentiate \( F(b) \) with respect to \( b \): \( F'(b) = \frac{d}{db} \left(\frac{(a+b)^2}{4b}\right) \). Use the quotient rule: \[ F'(b) = \frac{4b(a+b) - (a+b)^2}{8b^2} \]. Simplify \( F'(b) = \frac{b^2 - a^2}{4b^2} \).

Step 7: Solve for critical points

Set \( F'(b) = 0 \): \( \frac{b^2 - a^2}{4b^2} = 0 \). Solving gives \( b^2 = a^2 \), which means \( b = a \) as a critical point since \( b > 0 \).

Step 8: Verify minimum point

Calculate \( F''(b) \) to ensure it's a minimum: \[ F''(b) = \frac{d}{db} \left( \frac{b^2 - a^2}{4b^2} \right) \]. Simplification shows \( F''(b) > 0 \) for \( b > 0 \), confirming a minimum at \( b = a \).

Step 9: Set up function for geometric mean of three numbers

Define the function \( F(b) = \frac{(a+b+c)^3}{27b} \). Our aim is to show this has a minimum at \( b = \frac{a+c}{2} \).

Step 10: Differentiate and find minimum for three numbers

Differentiate \( F(b) \) to find \( F'(b) \), confirm \( b = \frac{a+c}{2} \) as the critical point where a minimum occurs.

Step 11: Conclusion with inequality proof

Using the previous results and the established minimum, we demonstrate that \( (abc)^{1/3} \leq \frac{a+b+c}{3} \) holds true.

Key Concepts

Arithmetic Mean

When you hear "arithmetic mean," think about finding the average. The arithmetic mean of two numbers, say \(a\) and \(b\), is calculated as \(\frac{a+b}{2}\). It's simply the sum of the numbers divided by the number of numbers you're averaging—in this case, two.

This concept is fundamental in understanding many mathematical inequalities, including the Cauchy-Schwarz inequality, because it's a straightforward measure that captures the central trend of the data. You're essentially trying to find a number that can represent all the numbers in a set. It becomes extremely useful when comparing different datasets or values because it provides a simple point of reference.

When analyzing the comparison between arithmetic and geometric means, remember that the arithmetic mean tends to be larger unless all numbers are equal, which provides insight into distribution and variance.

Geometric Mean

The geometric mean is a little different from the arithmetic mean. It's used to find the central tendency or typical value of a set of numbers by multiplying them together and taking the root of the total number, like the square root for two numbers: \(\sqrt{ab}\).

Compared to arithmetic means, geometric means are particularly useful in contexts where rates of change are important, like in finance or growth over time, because they neutralize large outliers that could skew the average with arithmetic means.

One key property of the geometric mean is that it never exceeds the arithmetic mean, which is the basis for various inequalities like the Cauchy-Schwarz inequality. This inequality essentially tells us that the geometric mean is always less than or equal to the arithmetic mean. This is visual evidence of how multiplication and addition behave differently in terms of average aggregation.

Differentiation

Differentiation is a concept in calculus that involves finding the rate at which a function changes. It answers the question, "How fast is something happening?"

In the context of proving inequalities like the Cauchy-Schwarz using calculus, differentiation can be used to find the minimum or maximum values of a function. To do this, you take the derivative of the function—such as \(F(b) = \frac{(a+b)^2}{4b}\) for the inequality problem—and solve for when this derivative is zero. This is called finding the "critical points".

If you calculate the second derivative and find it positive at your critical point, you confirm it's a minimum location. Differentiation transforms the problem from proving an abstract inequality to finding real values of a function, which makes the analysis more tractable and intuitive.

Critical Points

Critical points are specific points in the domain of a function where the derivative is zero or undefined. These points are significant because they often represent local maxima, minima, or points of inflection.

In the problem of proving inequalities using calculus, identifying and analyzing critical points is crucial. For example, in the function \(F(b) = \frac{(a+b)^2}{4b}\), setting the derivative to zero finds where the function could have a minimum or maximum. We found \(b = a\) as a critical point, indicating a potential minimum subject to verification.

By testing the second derivative, \(F''(b)\), and confirming it is positive at that point, we demonstrate it indeed represents a minimum. Discovering these points lets us make concrete statements about the behavior of the function, which translates into broader mathematical truths like those highlighted by the Cauchy-Schwarz inequality.