91Ó°ÊÓ

In mathematical induction, the base case is the starting point. It is where we verify that our statement or inequality is true for the initial value, typically for the smallest positive integer.
For our specific problem, we start by proving the inequality for \(n = 1\).
When \(n = 1\), our inequality becomes:
\[ 1 > 2(\sqrt{1+1} - 1)\] Simplifying the right side:
\[2 \times (\sqrt{2} - 1) \approx 0.828\] Since 1 is indeed greater than approximately 0.828, the inequality holds true for the base case. It means we can move to the next step with confidence that our base case is correct.

After establishing the base case, the next step in mathematical induction is to assume that our statement or inequality is true for some positive integer \(k\).
This assumption is what we call the inductive hypothesis.
For our problem, we assume that:
\[1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \text{...} + \frac{1}{\sqrt{k}} > 2(\sqrt{k+1} - 1)\]
This step is crucial because it sets up the framework for us to prove that if this assumption holds for the positive integer \(k\) (inductive hypothesis), then it should also hold for the next integer \(k+1\). This bridging step links the base case to all subsequent positive integers.

The inductive step is the heart of mathematical induction. Here, we use the inductive hypothesis to prove the statement for \(k + 1\).
For our problem, we need to show that:
\[1 + \frac{1}{\sqrt{2}} + \text{...} + \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k+1}} > 2(\sqrt{k+2} - 1)\]
We start by using our inductive hypothesis:
\[1 + \frac{1}{\sqrt{2}} + \text{...} + \frac{1}{\sqrt{k}} > 2(\sqrt{k+1} - 1)\] Then, add \(\frac{1}{\sqrt{k+1}}\) to both sides:
\[ 1 + \frac{1}{\sqrt{2}} + \text{...} + \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k+1}} > 2(\sqrt{k+1} - 1) + \frac{1}{\sqrt{k+1}} \]
By simplifying the right side, we focus on showing whether the left side surpasses the required expression, establishing that our inequality is maintained when extended to \(k+1\).
Combining terms and comparing expressions helps solidify this step.

Throughout mathematical induction, especially when dealing with inequalities, precision is key.
We constantly compare quantities to ensure one side is greater than or less than the other based on our hypotheses and logical progression.
In our example, inequality manipulation and simplification verify that our statement holds.
Simplifying inequalities often involves breaking them down into smaller steps or components that we already understand and proving each part logically.

• Always analyze both sides carefully.
• Use established mathematical rules for simplifying and comparing.

By ensuring each part maintains the inequality correctly, we solidify our mathematical proof. The principle of mathematical induction lets us extend truth from the smallest value to any positive integer, supported meticulously at each step.