Proof of Inequality (1+a)^n >= 1+na for a>-1 & n

In summary, Easty was close to proving an inequality that required na^2 to always be >= 0, but was unable to get there. He was helped by another user who provided a proof that na^2 always equals 0.
  • #1
Easty
19
0

Homework Statement



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

Homework Equations





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 don't 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
 
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  • #2
(1+ka)+a=1+(k+1)a,thus the inequality holds for n=k+1
 
  • #3
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
lanedance said:
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?

i had gotten to that stage, but thought it led no where.
The only term that isn't part of the original equality is na^2, since a is >-1, a^2 >= 0, so couldn't the extra term add some arbitary large postive number such that the inequality wasnt true?
 
  • #5
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
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
 

1. What is the proof of the inequality (1+a)^n >= 1+na for a>-1 & n?

The proof of this inequality is based on mathematical induction. First, we assume that it is true for n=1, i.e. (1+a)^1 >= 1+a. Then, we assume that it is true for n=k, i.e. (1+a)^k >= 1+ka. Finally, we prove that it is also true for n=k+1, which completes the induction step and proves the inequality for all values of n.

2. What is the significance of the inequality (1+a)^n >= 1+na for a>-1 & n?

This inequality is significant because it allows us to prove various other inequalities and mathematical theorems. It is also commonly used in calculus and analysis to prove the convergence of series and the continuity of functions.

3. Can you provide an example of how the inequality (1+a)^n >= 1+na is used in real-world applications?

One example of how this inequality is used in real-world applications is in the study of compound interest. For instance, if we deposit $100 in a bank account with an annual interest rate of 5%, the amount of money in our account after n years would be (1+0.05)^n * 100. By using the inequality, we can prove that this amount will always be greater than or equal to 100+0.05n, which represents the amount if we had earned simple interest.

4. Is the inequality (1+a)^n >= 1+na true for all values of a and n?

No, the inequality is only true for a>-1 and n=1,2,3,... In other words, it is only true for non-negative values of a and positive integer values of n. For negative values of a, the inequality may not hold.

5. How does the inequality (1+a)^n >= 1+na relate to other mathematical concepts?

This inequality is closely related to the binomial theorem, which states that (1+a)^n = 1+na+ n(n-1)a^2/2! + n(n-1)(n-2)a^3/3! + ... + a^n. The inequality can be seen as a simplified version of the binomial theorem, where we only consider the first two terms of the expansion.

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