analysis question

How do you show that (1-x)^n>=1-nx for n a natural number and x a real number? 

Have you ever heard of induction before?

1 clearly works.  So, assume true for n.  I tried
[
(1-(n+1)x)/(1-x)^(n+1)=1/(1-x)*[(1-nx)/(1-x)^n-x/(1-x)^n]
\]

Damn, stuck again.  ^_^

It follows immediately from the binomial theorem.

>>4 is half right, the binomial theorem makes it easy in the x<0 case.  (expand (1-x)^n then 1 and -nx will be your first two terms, since every term is positive when x<0, it ought to be strictly greater)  However, I fear the OP may have mislead us as the initial proposition doesn't seem to always hold true in the case x>0, example: x=4, n=3: (1-4)^3=>1-4*3 -> -27>=-11

OP here.  Check out Theorem 1 here:

http://planetmath.org/encyclopedia/11nnIsAnIncreasingSequence.html

I need to show what this theorem is proving, but I don't know about the "Since (1-x)^n>=1-nx, we have" part.  They do not show it, so to me, the proof is somewhat lacking. 

Here, they use substitute n-1 in for x, where n-1>=0 for n=1,2,...
 

I believe the inequality merely reverses when n is odd for n ≥ 3.

>>7, no, for sufficiently small, but still positive, x the relation should still hold.  Also, could you please explain what use an inequality is if the range on which it is valid is poorly defined, it's like saying that I have to values, one of which is either larger, smaller, or equal to the other: no useful information.

>>6

No, they substitute 1/n^2 for x.  which is sufficiently small, but still positive, for large values of n.