TITLE: demonstration by induction: $(1+a)^n ≥1+an$
QUESTION [6 upvotes]: Demonstrate by induction: $(1+a)^n ≥1+an$ is true, given a real number $a$, for any $n ∈ \mathbb N$. With $a > 0$
I need to demostre this using the induction principle. My doubt is in the second part of the demonstration.
This is what I have:
In order to demonstrate the predicate these two points must be true:

$P(1)$ must be True.
$∀ n ∈ \mathbb N , P(n) => P(n+1)$ is true.

We demonstrate the first point:
$(1+a)^1 ≥ 1+a*1$, 
$1+a ≥ 1+a$  So it is true
Now, the second part is where I have the problem. I do not know what to do. I understand the theory but I don't know how to apply it. 
I tried this:
$(1+a)^{n+1} ≥ 1+a(n+1)$    
But I don't see that as useful. 
Any tips?

REPLY [2 votes]: Before you do anything else, ask yourself: do you even believe that this is true?  Have you tried any examples with specific numbers?  Fixing the value of $a$ (especially taking $a$ to be very small) and varying the value of $n$, perhaps?  Does the result seem plausible after doing this?  Do your experiments suggest anything about what's going on?