# Proving Bernoulli's Inequality

1. Jul 22, 2013

### Seydlitz

1. The problem statement, all variables and given/known data
Prove Bernoulli's Inequality: if $h>-1$
$(1+h)^n \geq 1+hn$

2. Relevant equations
Binomial Theorem
$(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}$

3. The attempt at a solution
If $h=0$
$(1+0)^n=1$
$1=1$

If $h>0$
This
$(1+h)^n \geq 1+hn$
Implies
$(1+h)^n=\sum_{k}^{n}\binom{n}{k}h^{k}$
$\sum_{k=2}^{n}\binom{n}{k}h^{k} \geq 0$

So the proof is done.

2. Jul 22, 2013

### micromass

Staff Emeritus
And what if $h$ is negative?

3. Jul 22, 2013

### Seydlitz

Owh, I keep forgetting that the number after -1 is not 0.

$h>-1$ is equivalent to $h+1>0$

We know $\sum_{k=2}^{n}\binom{n}{k}h^{k} \geq 0$

Under the closure of multiplication positive number will always give positive number so:
$(h+1)\sum_{k=2}^{n}\binom{n}{k}h^{k} \geq 0$

$$\sum_{k=2}^{n}\binom{n}{k}h^{k+1}+\sum_{k=2}^{n}\binom{n}{k}h^{k}\geq 0$$

Is the reasoning fine?

4. Jul 22, 2013

### micromass

Staff Emeritus
I am not convinced of this if $h$ is negative. It requires a proof.

5. Jul 22, 2013

### Seydlitz

Ok will proof by induction works in this case, if there's negative number in play?

6. Jul 22, 2013

### micromass

Staff Emeritus
Induction on $n$ will work. Not sure why you think $h$ being negative will make an induction proof invalid.

7. Jul 22, 2013

### Seydlitz

Ahh, I didn't see that. I'm just a bit confused perhaps.

So the plan:
1.Proof the inequality is true using n=1
2.Assume n is true, to show n+1 is also true.

$(1+h)^{n+1}=(1+h)(1+h)^n$
$(1+h)(1+h)^n=(1+h)(1+nh)$
$(1+h)(1+nh)=1+nh+h+nh^2=1+(n+1)h+nh^2$
By inspection this should be true $1+(n+1)h+nh^2 \gek 1+(n+1)h$ for any h, and so the Bernoulli's inequality in consequence.

Last edited: Jul 22, 2013
8. Jul 22, 2013

### micromass

Staff Emeritus
This equality isn't true.

9. Jul 22, 2013

### Seydlitz

Ok I think I know where that goes wrong.

$(1+h)^{n+1}=(1+h)(1+h)^n$
$(1+h)(1+h)^n \geq (1+h)(1+nh)$ assuming the equality holds with $n$
$(1+h)(1+nh)=1+nh+h+nh^2=1+(n+1)h+nh^2$
$1+(n+1)h+nh^2 \geq 1+(n+1)h$
then
$(1+h)^{n+1} \geq 1+(n+1)h$

This is true right?
$1+(n+1)h+nh^2 \geq 1+(n+1)h$
$nh^2 \geq 0$

10. Jul 22, 2013

### micromass

Staff Emeritus
OK, that seems right.

11. Jul 22, 2013

### Seydlitz

Finally! Thank you for guiding me and clearing all of my messy mistakes. :D

Next question.