# Proof by induction

1. Jun 22, 2009

### Easty

1. The problem statement, all variables and given/known data

Using mathematical induction, prove the inequality
(1+a)^n >= 1+na for all a>-1 and all n

2. Relevant equations

3. The attempt at a solution
Base case
1+a >= 1+a. the inequality holds.

I am struggling with the inductive step.
by working backward, i multiply both sides by (1+a) and this gives me

(1+a)^(k+1) >= (1+ka)(1+a)

all i can think of at this point is that 1+a >= 0, but i dont think this helps me.

working the other way, starting with n=k+1, i find

(1+a)(1+a)^k>= (1+ka)+a

would is be enough to state that (1+a)> a and thus the inequality holds?

thanks in advance

2. Jun 22, 2009

### kof9595995

(1+ka)+a=1+(k+1)a,thus the inequality holds for n=k+1

3. Jun 22, 2009

### lanedance

Hey Easty, i think you were close with your first step, not to sure about your 2nd,

how about this, so as usual start by assuming the case for n is true and trying to show n+1 is also true based on this:

(1+a)^(n+1) = (1+a)*(1+a)^n >= (1+a)*(1+na) = (1 + a + na + na^2) = 1 + (n+1)a + na^2

So now you have
(1+a)^(n+1) >= 1 + (n+1)a + na^2

any ideas from here?

4. Jun 22, 2009

### Easty

i had gotten to that stage, but thought it led no where.
The only term that isnt part of the original equality is na^2, since a is >-1, a^2 >= 0, so couldnt the extra term add some arbitary large postive number such that the inequality wasnt true?

5. Jun 22, 2009

### lanedance

so if na^2 is always >=0 can you think of another inequality to write that will help get you close to the required form?

6. Jun 23, 2009

### Easty

I think i've got it
na^2 >= 0

add 1+(n+1)a to both sides

1+(n+1)a+na^2 >= 1+(n+1)a

therefore

(1+a)^(n+1) >= 1 + (n+1)a + na^2 >= 1 + (n+1)a

(1+a)^(n+1) >=1 + (n+1)a as required

i'm pretty sure this is correct. Thanks for all your help lanedance

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