Inequality induction proof

In summary: AM-GM inequality, the denominator is greater than 2, and you get the desired result.In summary, the conversation involves proving the inequality 2(\sqrt{n+1}-\sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n}-\sqrt{n-1}) for n \in N. Different approaches are discussed, including a direct proof and a proof by induction, with the direct proof being deemed easier. The AM-GM inequality is used to prove the desired result.
  • #1
Boombaard
10
0

Homework Statement


Show that [tex] 2( \sqrt{n+1} - \sqrt{n} ) < \frac{1}{ \sqrt{n}} < 2( \sqrt{n} - \sqrt{n-1}), n \in N [/tex]

2. The attempt at a solution

this works for n=1, but I'm entirely at a loss when trying to figure out which equations i can assume to be 'obviously' true (for the sake of the induction step), and what cannot be assumed..
All sides of the inequality eventually become 0, and while it seems fairly obvious to me that they all do it at an incrementally slower pace than the preceding one, I'm not sure what to make of this mathematically, or how to mold it into an equation expressing that.. can anyone lend a hand?

@radou: sorry, took a few mins to figure out what i'd done wrong
 
Last edited:
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  • #2
You may want to correct your LaTeX code.
 
  • #3
Unless the question says "prove by induction", a direct proof is easy.

Take each inequality separately, multiply both sides by \root n, and it should be fairly obvious what to do after that.

Another direct proof: think about approximations to
[tex] \int x^{-1/2} \, dx [/tex]
 
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  • #4
I can't possibly see how induction could do it.

Daniel.
 
  • #5
indeed, a direct proof is much much easier than induction... the inequalities are equivalent to:
[tex]\sqrt{n+1}-\sqrt{n}>0[/tex]
 
  • #6
I suppose you could argue that a direct proof IS a proof by induction.

You say "assume the result is true for n = k-1", then show it is true for n = k ... even though you never actually used the assumption :wink:
 
  • #7
bah. i feel like I'm really not very bright here, but why is the difference between the sqrts of 2 successive natural numbers always < .5 ?
 
  • #8
multiply out:

[tex](\sqrt{n+1}-\sqrt{n}) \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}[/tex]
 

1. What is inequality induction proof?

Inequality induction proof is a method commonly used in mathematical and scientific fields to prove that a certain inequality holds true for all values of a variable. It involves showing that the inequality is true for the smallest possible value of the variable, and then using mathematical reasoning to show that it must also be true for the next value, and so on.

2. How is inequality induction proof different from other methods of proof?

Unlike other methods of proof, such as direct proof or proof by contradiction, inequality induction proof does not require a specific starting point or assumption. Instead, it relies on the concept of a variable and its relationship to the inequality being proven.

3. When is inequality induction proof used?

Inequality induction proof is commonly used when trying to prove a mathematical or scientific statement that involves a variable, such as showing that a certain equation or inequality holds true for all values of the variable.

4. What are the steps involved in inequality induction proof?

The steps involved in inequality induction proof typically include: (1) showing that the inequality holds true for the smallest possible value of the variable, (2) assuming that the inequality is true for a specific value of the variable, (3) using mathematical reasoning to show that it is also true for the next value of the variable, and (4) repeating this process until the inequality is proven for all values of the variable.

5. What are the limitations of inequality induction proof?

Inequality induction proof can only be used to prove inequalities involving a single variable. It also requires a certain level of mathematical knowledge and understanding to use effectively. Additionally, it may not always be the most efficient method of proof for certain statements.

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