# Inequality induction proof

1. Jan 10, 2007

### Boombaard

1. The problem statement, all variables and given/known data
Show that $$2( \sqrt{n+1} - \sqrt{n} ) < \frac{1}{ \sqrt{n}} < 2( \sqrt{n} - \sqrt{n-1}), n \in N$$

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: Jan 10, 2007
2. Jan 10, 2007

You may want to correct your LaTeX code.

3. Jan 10, 2007

### AlephZero

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
$$\int x^{-1/2} \, dx$$

Last edited: Jan 10, 2007
4. Jan 11, 2007

### dextercioby

I can't possibly see how induction could do it.

Daniel.

5. Jan 11, 2007

### tim_lou

indeed, a direct proof is much much easier than induction... the inequalities are equivalent to:
$$\sqrt{n+1}-\sqrt{n}>0$$

6. Jan 12, 2007

### AlephZero

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

7. Jan 14, 2007

### Boombaard

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. Jan 15, 2007

### matt grime

multiply out:

$$(\sqrt{n+1}-\sqrt{n}) \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}$$