Is This Induction Proof of Bernoulli's Inequality Correct?

[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

 

[Muzza]

No.

Induction step:
A(n) is true is since n = 1...

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

 

[Bob19]

Okay and thank you for your answer,
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

 

[Muzza]

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

 

[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.

 

[Bob19]

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.

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

 

[VietDao29]

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}
Bob

?
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.