# Bernoulli inequality proof

1. Nov 9, 2005

### Bob19

Hi I'm doing a small induction proof for bernoullis inequailty:
Proof:
Given the inequality $$A(n) = (1+x) ^n \geq 1+nx$$
$$r \geq -1$$, $$n \in \mathbb{N}$$
Initial step:
A(n=1) is true cause $$(1+x) \geq 1 + x$$ is true.
Induction step:
A(n) is true is since n = 1 and $$r \geq -1$$ so
$$0 \geq 0$$
Therefore by the rules of induction
A(n+1) is true.
q.e.d.
Is my proof sufficient ??
Best Regards,
Bob

Last edited: Nov 9, 2005
2. Nov 9, 2005

### Muzza

No.

In the inductive step, n is supposed to be an /arbitrary/ natural number for which A(n) is true.

3. Nov 9, 2005

### Bob19

What do I need to add under the induction step to complete the proof?
Do I need to show that A(n+1) is true?
By that I mean

$$A(n+1) = (1+x) ^{n} (1+x) \geq (1 + nx) (1+x)$$
$$A(n+1) = (1+x) ^{n} (1+x) \geq (1 + x + nx^2)$$

Is this correct to way to show that A(n+1) is true ?
Sincerley
Bob

Last edited: Nov 9, 2005
4. Nov 9, 2005

### Muzza

Yes. Given that A(n) is true for some (arbitrary) n, show that A(n + 1) is true.

5. Nov 9, 2005

### amcavoy

You could also prove it by saying:

$$f\left(x\right)=\left(1+x\right)^{n}-1-nx,$$

then use derivatives to show that it is greater than zero and increasing on a certain interval.

6. Nov 9, 2005

### Bob19

Correct me if I'm wrong:

d/dx = n x ^ (n-1)

Since $$n \in \mathbb{N}$$ then if n > 0, A(n+1) groves and therefore its true.

Is that what you mean ?

best regards,
Bob

7. Nov 10, 2005

### VietDao29

???
I just wondered what r is in your first post??
-----------------
There are 3 steps in proof by induction:
(1) Test if the statement's true for n = 0.
(2) Assume the statement is true for n = k.
(3) Prove the statement is true for n = k + 1 using the induction hypothesis (2).
----------
(1) So you have shown that for n = 1, the equality is true. Or you can even show that the inequality is true for n = 0.
For n = 0, you'll have:
(1 + x)0 ≥ 1 + 0x. And that's true!
(2)Then assume the inequality is true for n = k, ie:
(1 + x)k ≥ 1 + kx.
(3)Now let's prove it's true for n = k + 1. That means, you have to prove:
(1 + x)k + 1 ≥ 1 + (k + 1)x.
So (1 + x)k + 1 = (1 + x)k (1 + x)
Since x ≥ -1, so 1 + x ≥ 0.
Using the induction hypothesis, you have:
(1 + x)k ≥ 1 + kx
<=> (1 + x)k (1 + x) ≥ (1 + kx) (1 + x) (Note that: 1 + x ≥ 0).
<=> (1 + x)k (1 + x) ≥ 1 + x + kx + kx2. From there, can you show that (1 + x)k + 1 ≥ 1 + (k + 1)x?
---------------------
You don't have to prove it by taking the derivatives of f(x). But by the way, your f'(x) is wrong.
f(x) = (1 + x)n - 1 - nx, then
f'(x) = (1 + x)n' - 1' - nx' = n(1 + x)n - 1 - n.

Last edited: Nov 10, 2005