Proof of Inequality by Binomial Theorem I have been stuck in this proof for a while. This should not be a long shot.
Claim If $x_n>0$ for all $n\in N$, then $1+nx_n\leq (1+x_n)^n$.
My strategy is to use Induction Theorem and Binomial Theorem, which does not work quite well.
My proof If $n=1$, $1+x_1\leq (1+x_1)^1$ (True).
Assume this claim is true for $n$. Consider $n+1$:
$(1+x_{n+1})^{n+1}=x_{n+1}^{n+1}+(n+1)x_{n+1}^n+\dots+2x_{n+1}+1$. (**)
Need to show $(**)\geq 1+(n+1)x_{n+1}$, given $(1+nx_n)\leq (1+x_n)^n$, i.e. $(1+nx_n)\leq x_n^n+nx_n^{n-1}+\dots + 2x_n+1$.
Could someone give me a hint on how to deal with the $n+1$ case? In particular, how to link $x_n$ to $x_{n+1}$, if induction is the right way to go? Thanks in advance!
 A: Note that the induction doesn't work here because you have not a relation between $x_n$ and $x_{n+1}$. You need to prove the inequality for every fixed $n$.
A: The way I look at this one is to temporarily forget about the index $n$ on the $x_n$ and simply note that, for any real $y > 0$ and $1 \le n \in \Bbb N$,
$1 + y \le 1 + ny \le \sum_0^n \dfrac{n!}{i!(n - i)!} 1^{n - i} y^i = (1 + y)^n, \tag 1$
which follows from the binomial theorem applied to $(1 + y)^n$; since (1) holds for every $n$, we may successively set $y = x_n$ for any positive real sequence $x_n$ and obtain
$1 + x_n \le (1 + x_n)^n \tag 2$
for each $n$.
The fact is that the problem specifies no relationship 'twixt $x_n$ and $x_{n + 1}$, etc., so attempting to use one as a means of solution is ill-founded; the $x_n$ are arbitrary and independent.
So it's really all about (1).  The essential relaitionship
$1 + y \le (1 + y)^n \tag 3$
can in fact be proved by induction without resorting to the binomial theorem:  if 
$1 + y \le (1 + y)^k, \tag 4$
then since
$1 < 1 + y, \tag 5$
we have
$1 + y < (1 + y)(1 + y) \le (1 + y)^k (1 + y) = (1 + y)^{k + 1}. \tag 6$
"About binomial theorem I am teeming with a lot of news,
With many cheerful facts about the square of the hypoteneuse!"
--Gilbert & Sullivan, "Modern Major General".
