I'm asked to used induction to prove Bernoulli's Inequality: If $1+x>0$, then $(1+x)^n\geq 1+nx$ for all $n\in\mathbb{N}$. This what I have so far:
Let $n=1$. Then $1+x\geq 1+x$. This is true. Now assume that the proposed inequality holds for some arbitrary $k$, namely that
$$1+x>0\implies (1+x)^k\geq 1+kx,~\forall~k\in\mathbb{N}\setminus\{1\}$$
is true. We want to show that the proposed inequality holds for $k+1$. Thus multiplication of $(1+x)$ on each side of the above inequality produces the following result:
$$(1+x)(1+x)^k\geq (1+kx)(1+x)\implies (1+x)^{k+1}\geq 1+x+kx+kx^2\cdots\cdots\cdots$$
I'm not sure where to go from here.
Best Answer
You're almost done. $(1+x)^{k+1}\geq (1+kx)(1+x)=1+kx^2+(k+1)x\geq1+(k+1)x$ since $kx^2\geq 0$.
Also in your assumption, "$1+x>0\implies (1+x)^k\geq 1+kx,~\forall~k\in\mathbb{N}\setminus\{1\}$", you shouldn't write $\forall~k\in\mathbb{N}\setminus\{1\}$. Because if you assume this is true for all $k\in \mathbb{N}$, then you've already assumed it is true for $k+1$. Also, you shouldn't let $k\ne 1$, since then your argument doesn't allow you to conclude $P(1) \implies P(2)$.