Problem 20
Question
Let \(a>0 .\) Use mathematical induction to prove that $$ \sqrt{a+\sqrt{a+\sqrt{a+\cdots+\sqrt{a}}}}<\frac{1+\sqrt{4 a+1}}{2} $$ where the left member contains an arbitrary number of radicals.
Step-by-Step Solution
Verified Answer
The inequality holds for any number of nested radicals using induction.
1Step 1: Understanding the Problem
We need to prove the inequality \( \sqrt{a + \sqrt{a + \sqrt{a + \cdots + \sqrt{a}}}} < \frac{1 + \sqrt{4a + 1}}{2} \), where the left side has repeated nested radicals an arbitrary number of times.
2Step 2: Establish the Base Case
Let's denote the repeated radical as a sequence: \( x_1 = \sqrt{a} \). We'll prove the base case: \( x_1 < \frac{1 + \sqrt{4a + 1}}{2} \). Calculate \( \sqrt{a} < \frac{1 + \sqrt{4a+1}}{2} \) by rewriting as \( \sqrt{4a + 1} > 2\sqrt{a} - 1 \). This holds as both sides are positive.
3Step 3: Inductive Hypothesis
Assume for some \( k \), \( x_k = \sqrt{a + x_{k-1}} < \frac{1 + \sqrt{4a+1}}{2} \) holds true. This is our inductive hypothesis.
4Step 4: Inductive Step
We need to show \( x_{k+1} < \frac{1 + \sqrt{4a+1}}{2} \). Consider \( x_{k+1} = \sqrt{a + x_k} \), then \( \sqrt{a + x_k} < \frac{1 + \sqrt{4a+1}}{2} \). By squaring both sides and using the hypothesis, verify \( a + x_k < \left( \frac{1+\sqrt{4a+1}}{2} \right)^2 \).
5Step 5: Conclusion
Using the principles of mathematical induction, we have shown the inequality holds for the base case and holds from \( x_k \) to \( x_{k+1} \). Thus the inequality \( \sqrt{a + \sqrt{a + \cdots + \sqrt{a}}} < \frac{1 + \sqrt{4a + 1}}{2} \) holds for any arbitrary number of nested radicals.
Key Concepts
Nested RadicalsInequality ProofInductive Hypothesis
Nested Radicals
Nested radicals involve a sequence where the operation of taking the square root is applied repeatedly in a nested manner. This can look overwhelming at first, but it's quite simple once you break it down. For example, in the expression \( \sqrt{a + \sqrt{a + \cdots + \sqrt{a}}} \), you have a sequence of square roots, each on top of the earlier one.
These nested expressions often converge to a particular value as more terms are added. Here, our task involves demonstrating that regardless of how many radicals you nest, the expression remains under a certain boundary, represented by \( \frac{1 + \sqrt{4a + 1}}{2} \).
Nested radicals often appear in problems involving limits and are common in mathematical series or sequences. Visualizing them as sequences can simplify understanding their behavior.
These nested expressions often converge to a particular value as more terms are added. Here, our task involves demonstrating that regardless of how many radicals you nest, the expression remains under a certain boundary, represented by \( \frac{1 + \sqrt{4a + 1}}{2} \).
Nested radicals often appear in problems involving limits and are common in mathematical series or sequences. Visualizing them as sequences can simplify understanding their behavior.
Inequality Proof
Proving inequalities is a vital skill in mathematics. It involves demonstrating that one expression is always greater or lesser than another. In this problem, we are tasked with proving that a nested radical expression is always less than a given value.
The key to inequality proofs like this one is starting with smaller cases or easier scenarios, then building a broader argument that applies generally. This specific proof uses a defined sequence and shows step by step that the inequality holds for each element of the sequence.
To begin, verifying the base case, which involves the simplest form, can set the tone. From there, you generalize it by assumption and using properties of numbers like the transitive property to extend the inequality step to step.
The key to inequality proofs like this one is starting with smaller cases or easier scenarios, then building a broader argument that applies generally. This specific proof uses a defined sequence and shows step by step that the inequality holds for each element of the sequence.
To begin, verifying the base case, which involves the simplest form, can set the tone. From there, you generalize it by assumption and using properties of numbers like the transitive property to extend the inequality step to step.
Inductive Hypothesis
An inductive hypothesis is an essential component of a mathematical induction proof. It involves assuming that a statement holds for some arbitrary integer \( k \), and using that assumption to prove it holds for \( k+1 \).
In our inequality proof, the inductive hypothesis is that our inequality \( x_k = \sqrt{a + x_{k-1}} < \frac{1 + \sqrt{4a + 1}}{2} \) is valid for a specific \( k \).
The power of the inductive hypothesis lies in its ability to bridge the proof from a specific base case to general applicability. Once the base case is proven, the inductive hypothesis allows us to step through each consecutive case, thus concluding that the statement is true for all natural numbers.
In our inequality proof, the inductive hypothesis is that our inequality \( x_k = \sqrt{a + x_{k-1}} < \frac{1 + \sqrt{4a + 1}}{2} \) is valid for a specific \( k \).
The power of the inductive hypothesis lies in its ability to bridge the proof from a specific base case to general applicability. Once the base case is proven, the inductive hypothesis allows us to step through each consecutive case, thus concluding that the statement is true for all natural numbers.
Other exercises in this chapter
Problem 19
(Nesbitt's Inequality) Let \(a, b, c\) be strictly positive real mumbers. Then $$ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geq \frac{3}{2} $$
View solution Problem 19
Let \(f^{\mid 1]}=f\) be given by \(f(x)=\frac{1}{1-x} .\) Find (i) \(f^{\mid 2]}(x)=(f \circ f)(x)\) (ii) \(f^{[3]}(x)=(f \circ f \circ f)(x)\), and \((x)\) (i
View solution Problem 20
Let \(f: A \rightarrow B\) and \(g: B \rightarrow C\) be functions. Shew that (i) if \(g \circ f\) is injective, then \(f\) is injective. (ii) if \(g \circ f\)
View solution Problem 21
Let \(a, b, c\) be positive real numbers. Prove that $$ (a+b)(b+c)(c+a) \geq 8 a b c . $$
View solution