Problem 85 - Calculus with Concepts in Calculus

Question

Prove that if \(00 .\right)\)

Step-by-Step Solution

Answer

Thus, if \(0

Step 1: Understand the problem

We need to prove that if two positive numbers are such that \ a < b, then the geometric mean \ \( \sqrt{ab} \) is greater than \( a \) and less than the arithmetic mean \( \frac{a+b}{2} \).

Step 2: Prove \(a < \sqrt{ab}\)

Since \(0 < a < b\), we know that \(b > a\). Therefore, \ ab > a^2 \, (since multiplying both sides by \(a > 0\)). Taking square roots gives \(\sqrt{ab} > \sqrt{a^2} = a\).

Step 3: Introduce the hint

Let us use the hint, which involves the expression \((\sqrt{b/2} - \sqrt{a/2})^2 > 0\). Since it is the square of a real number, it's always greater than or equal to 0, but we'll explore when it's strictly positive.

Step 4: Expand the quadratic expression

Expand \((\sqrt{b/2} - \sqrt{a/2})^2:\)\[ (\sqrt{b/2} - \sqrt{a/2})^2 = \left(\sqrt{\frac{b}{2}}\right)^2 - 2\sqrt{\frac{b}{2}}\sqrt{\frac{a}{2}} + \left(\sqrt{\frac{a}{2}}\right)^2 \] This simplifies to \[ \frac{b}{2} - \sqrt{ab} + \frac{a}{2} > 0 .\]

Step 5: Simplify inequality

Adding \(\sqrt{ab}\) to both sides yields \[ \frac{b}{2} + \frac{a}{2} > \sqrt{ab} .\] This can be rewritten as \( \frac{a+b}{2} > \sqrt{ab} \), demonstrating that the geometric mean is less than the arithmetic mean.

Step 6: Combine results

We have shown that \(a < \sqrt{ab}\) and \(\sqrt{ab} < \frac{a+b}{2}\). Together, they complete the proof of \(a < \sqrt{ab} < \frac{a+b}{2}\).

Key Concepts

Inequality Proof

Proving inequalities involves showing that one quantity is greater or less than another. In this exercise, we need to prove that the geometric mean of two positive numbers, \(a\) and \(b\), satisfies certain conditions. Specifically, when \(0 < a < b\), the geometric mean \(\sqrt{ab}\) will always be greater than \(a\) and less than their arithmetic mean \((a+b)/2\). To establish this inequality, we approach it in two parts:

• Show that \(a < \sqrt{ab}\).
• Show that \(\sqrt{ab} < \frac{a+b}{2}\).

Understanding these comparisons requires clear steps to prove each part based on the given conditions.

Square Roots

The concept of square roots is fundamental in our proof. A square root of a number \(x\) is a number that, when multiplied by itself, gives \(x\). In this exercise, we use square roots to manipulate the expression \(\sqrt{ab}\). This is the geometric mean of \(a\) and \(b\), indicating that it is the square root of their product.To prove \(a < \sqrt{ab}\), we start with the inequality \(ab > a^2\). This follows from the conditions that \(a < b\). By taking the square root of both sides, we obtain \(\sqrt{ab} > a\).Square roots help bridge products into comparable forms in inequalities, linking the geometric mean with other forms like the arithmetic mean.

Arithmetic Mean

The arithmetic mean of two numbers is the sum of the numbers divided by two. It represents their average. In this exercise, the key challenge is to demonstrate that the geometric mean \(\sqrt{ab}\) is less than the arithmetic mean \((a+b)/2\).To establish this part of the inequality, we utilize a hint concerning the square of the difference between the square roots of halves of \(a\) and \(b\). Specifically, the expression \((\sqrt{b/2} - \sqrt{a/2})^2 > 0\) is explored. By expanding this expression, we achieve:\[\left(\sqrt{\frac{b}{2}}\right)^2 - 2\sqrt{\frac{b}{2}}\sqrt{\frac{a}{2}} + \left(\sqrt{\frac{a}{2}}\right)^2 > 0\]This simplifies to \(\frac{b}{2} - \sqrt{ab} + \frac{a}{2} > 0\), leading to \(\frac{a+b}{2} > \sqrt{ab}\). Thus, we show that the arithmetic mean surpasses the geometric mean in this context.

Algebraic Manipulation

Algebraic manipulation is a powerful tool that allows us to rearrange and simplify expressions. This skill is crucial when working through mathematical proofs like the one in our exercise. To successfully prove the inequality, we perform several steps that involve rearranging terms and taking square roots.- In proving \(a < \sqrt{ab}\), we manipulate the inequality \(a < b\) to get \(ab > a^2\), then take square roots.- The hint provided involves expanding a square, which requires careful manipulation to make evident that the result is positive, ultimately leading to \(\frac{a+b}{2} > \sqrt{ab}\).Mastering algebraic manipulation involves understanding how to strategically handle terms to transform an expression into a desired form, simplifying and solving complex inequality problems.

Previous

Problem 84

Next

Problem 87

Other exercises in this chapter

Problem 83

Show that \(|a+b|=|a|+|b|\) if and only if \(a b \geq 0\) (which means that \(a=0, b=0\), or \(a\) and \(b\) have the same sign).

View solution

Problem 84

Prove that if \(a

View solution

Problem 87

Let \(0

View solution

Problem 88

Prove that \(\sqrt{2}\) is irrational. (Hint: Assume that \(\sqrt{2}=\) \(p / q\), where \(p\) and \(q\) are integers such that at most one of them is divisible

View solution

Key Concepts

Inequality Proof

Proving inequalities involves showing that one quantity is greater or less than another. In this exercise, we need to prove that the geometric mean of two positive numbers, \(a\) and \(b\), satisfies certain conditions. Specifically, when \(0 < a < b\), the geometric mean \(\sqrt{ab}\) will always be greater than \(a\) and less than their arithmetic mean \((a+b)/2\). To establish this inequality, we approach it in two parts:

• Show that \(a < \sqrt{ab}\).
• Show that \(\sqrt{ab} < \frac{a+b}{2}\).

Understanding these comparisons requires clear steps to prove each part based on the given conditions.

Square Roots

The concept of square roots is fundamental in our proof. A square root of a number \(x\) is a number that, when multiplied by itself, gives \(x\). In this exercise, we use square roots to manipulate the expression \(\sqrt{ab}\). This is the geometric mean of \(a\) and \(b\), indicating that it is the square root of their product.To prove \(a < \sqrt{ab}\), we start with the inequality \(ab > a^2\). This follows from the conditions that \(a < b\). By taking the square root of both sides, we obtain \(\sqrt{ab} > a\).Square roots help bridge products into comparable forms in inequalities, linking the geometric mean with other forms like the arithmetic mean.

Arithmetic Mean

The arithmetic mean of two numbers is the sum of the numbers divided by two. It represents their average. In this exercise, the key challenge is to demonstrate that the geometric mean \(\sqrt{ab}\) is less than the arithmetic mean \((a+b)/2\).To establish this part of the inequality, we utilize a hint concerning the square of the difference between the square roots of halves of \(a\) and \(b\). Specifically, the expression \((\sqrt{b/2} - \sqrt{a/2})^2 > 0\) is explored. By expanding this expression, we achieve:\[\left(\sqrt{\frac{b}{2}}\right)^2 - 2\sqrt{\frac{b}{2}}\sqrt{\frac{a}{2}} + \left(\sqrt{\frac{a}{2}}\right)^2 > 0\]This simplifies to \(\frac{b}{2} - \sqrt{ab} + \frac{a}{2} > 0\), leading to \(\frac{a+b}{2} > \sqrt{ab}\). Thus, we show that the arithmetic mean surpasses the geometric mean in this context.

Algebraic Manipulation

Algebraic manipulation is a powerful tool that allows us to rearrange and simplify expressions. This skill is crucial when working through mathematical proofs like the one in our exercise. To successfully prove the inequality, we perform several steps that involve rearranging terms and taking square roots.- In proving \(a < \sqrt{ab}\), we manipulate the inequality \(a < b\) to get \(ab > a^2\), then take square roots.- The hint provided involves expanding a square, which requires careful manipulation to make evident that the result is positive, ultimately leading to \(\frac{a+b}{2} > \sqrt{ab}\).Mastering algebraic manipulation involves understanding how to strategically handle terms to transform an expression into a desired form, simplifying and solving complex inequality problems.