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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!

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    $\begingroup$ $(1+x)^{n+1}=x^{n+1}+\cdots+(n+1)x+1$. $\endgroup$ – Lord Shark the Unknown Aug 1 '18 at 3:19
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    $\begingroup$ Try multiplying both sides of $(1+x)^n\ge1+nx$ by $1+x$. $\endgroup$ – Simply Beautiful Art Aug 1 '18 at 3:20
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    $\begingroup$ Just prove that $\forall x>0\;\forall n\in \Bbb N\;(1+nx\leq (1+x)^n))$ by induction on $n$. $\endgroup$ – DanielWainfleet Aug 1 '18 at 4:18
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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".

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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 $n$.

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