# Inequality proof

1. Feb 21, 2010

### jeff1evesque

1. The problem statement, all variables and given/known data
$$\frac{1}{2^{k}+1} + \frac{1}{2^{k} +2} + ... + \frac{1}{2^k + 2^k} \geq \frac{1}{2}$$

2. The attempt at a solution
Not too sure, I am working on a larger proof (not too much difficult) and the above is my attempt to prove the induction step k+1 (since $$\frac{1}{2^k + 2^k} = \frac{1}{2^{k+1}}$$).

Should i try to factor out $$\frac{1}{2^k}$$?

2. Feb 21, 2010

### tiny-tim

Hi jeff!

Hint: they're all greater than the last one.

3. Feb 21, 2010

### jeff1evesque

I still don't see it, i wish it were clear to me.

Thanks though,

Jeffrey Levesque

4. Feb 21, 2010

### l'Hôpital

If a < b < c
a + a + a < a + b + c

Right?

Try something along those lines.

5. Feb 21, 2010

### Dick

Don't you agree with tiny-tim that the last term in the sum is smaller than the rest? How many terms in the sequence are there?

6. Feb 21, 2010

### jeff1evesque

Yes I agree, but that isn't helping me at all- I just don't know how to formulate a proof for this.

7. Feb 21, 2010

### jeff1evesque

There are $$2^{k+1}$$ terms

8. Feb 21, 2010

### l'Hôpital

Are you sure? Count again. : )

9. Feb 21, 2010

### jeff1evesque

$$2^k$$ terms.

10. Feb 21, 2010

### jeff1evesque

never minnd, I actually did this in a different problem- thanks everyone

11. Feb 21, 2010

### snipez90

I also noticed that the last term on the LHS is the smallest, but I was confused for awhile since I thought the denominators were 2^k + 2^0, 2^k + 2^1, ..., 2^k + 2^k, in which case there are k+1 terms, and the estimate fails.

Anyways assuming the progression is what everyone else thinks it is, then basically $2^i \leq 2^k$ for i = 1, 2, ..., k so
$$2^k + 2^i \leq 2^k + 2^k = 2^{k+1} \Rightarrow \frac{1}{2^k + 2^i} \geq \frac{1}{2^{k+1}}$$
for i = 1, 2, ..., k.