Today, we’re diving into a bit of elementary number theory to prove a specific inequality. We aim to demonstrate that for positive integers n where 1 ≤ n ≤ 4, the inequality n² + 1 ≥ 2ⁿ holds true. This isn’t a general proof for all integers, mind you, just for this specific range! It’s a nice, concrete exercise that illustrates the concept of proof by cases.

Essentially, we will show this to be true by checking every single case in the given range. Let’s get started.

Case 1: n = 1

When n = 1, the left side of the inequality becomes:

1² + 1 = 1 + 1 = 2

The right side of the inequality is:

2¹ = 2

Thus, 2 ≥ 2, which is a true statement. The inequality holds for n = 1.

Case 2: n = 2

When n = 2, the left side of the inequality is:

2² + 1 = 4 + 1 = 5

The right side of the inequality is:

2² = 4

Thus, 5 ≥ 4, which is true. The inequality holds for n = 2.

Case 3: n = 3

When n = 3, the left side becomes:

3² + 1 = 9 + 1 = 10

The right side becomes:

2³ = 8

Thus, 10 ≥ 8, which is true. The inequality holds for n = 3.

Case 4: n = 4

When n = 4, the left side becomes:

4² + 1 = 16 + 1 = 17

The right side becomes:

2⁴ = 16

Thus, 17 ≥ 16, which is true. The inequality holds for n = 4.

Conclusion

We have shown, through a case-by-case analysis, that the inequality n² + 1 ≥ 2ⁿ holds true for n = 1, 2, 3, and 4. While this method is only viable for a finite number of cases, it’s a solid approach to demonstrate a statement within a specific range, as we just did. This simple example also demonstrates how a mathematical statement may hold true in a certain range but not generally (e.g., for n>4, this inequality does not hold). It is important to be specific in the limitations of your conclusions.

Stay tuned for more mathematical musings!