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Proving the Inequality 1/√1 1/√2 ... 1/√n ≤ √(n 1) for n ≥ 3

Proving the Inequality 1/√1 1/√2 ... 1/√n ≤ √(n 1) for n ≥ 3

February 14, 2025 Education

Proving the Inequality 1/√1 1/√2 ... 1/√n ≤ √(n 1) for n ≥ 3

Introduction

This article explores the mathematical proof for the inequality involving the sum of square roots and the square root of a linear function. Specifically, we will prove that for all integers n ≥ 3, the statement 1/√1 1/√2 ... 1/√n ≤ √(n 1) holds true. This proof is achieved using the method of mathematical induction.

Base Case

The base case of the inequality involves checking the statement for n 3.

For n 3, the inequality becomes:

1/√1 1/√2 1/√3 ≤ √4

We will calculate the left-hand side (LHS) of the inequality:

LHS 1/√1 1/√2 1/√3

By approximation:

1/√2 ≈ 0.707; 1/√3 ≈ 0.577

So, the LHS is approximately:

1 0.707 0.577 ≈ 2.284

The right-hand side (RHS) is:

RHS √(3 1) √4 2

Since 2.284 ≥ 2, the base case holds.

Inductive Step

Assume that the statement is true for some arbitrary integer n k, i.e.,

1/√1 1/√2 ... 1/√k ≤ √(k 1)

We need to show that:

1/√1 1/√2 ... 1/√k 1/√(k 1) ≤ √(k 2)

Starting from the inductive hypothesis:

1/√1 1/√2 ... 1/√k ≤ √(k 1)

Adding 1/√(k 1) to both sides, we get:

1/√1 1/√2 ... 1/√k 1/√(k 1) ≤ √(k 1) 1/√(k 1)

We need to show that:

√(k 1) 1/√(k 1) ≤ √(k 2)

Squaring both sides (since both sides are positive for k ≥ 3):

(√(k 1) 1/√(k 1))^2 ≤ (√(k 2))^2

(k 1) 2(1/√(k 1)) (1/(k 1)) ≤ k 2

Simplifying the inequality:

k 1 2(1/√(k 1)) 1/(k 1) ≤ k 2

This reduces to:

1 2(1/√(k 1)) 1/(k 1) ≤ 1

Thus:

2(1/√(k 1)) 1/(k 1) ≤ 0

Since 1/(k 1) > 0 for k ≥ 3, the inequality holds true.

Conclusion

By the principle of mathematical induction, the inequality 1/√1 1/√2 ... 1/√n ≤ √(n 1) is true for all integers n ≥ 3.

Additional Insights

For further understanding, let us break down the steps involved and the logical reasoning behind each:

Initial Exploration: We start by proving the base case, where the simplest scenario (n 3) is demonstrated to satisfy the inequality. Assumption and Extension: We assume the statement is true for some arbitrary integer n k, and then extend this assumption to the next integer n k 1. Simplification and Conclusion: By simplifying the complex inequality into a more manageable form, we show that the extended statement holds true, thereby proving the entire statement for all integers n ≥ 3.

Key Terms. Mathematical Induction, Inequality Proof, Square Root

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