For any $x \in \mathbb{R}$, $x > -1$, $(1 + x) ^ n \geq 1 + nx$ for all $n \in \mathbb{ N }$
I know the steps of induction, (base case, assume, prove), but the was this one is set up is completely different from any other problem I have seen. Having two variables in the same equation is what is confusing me, and I honestly have no idea where to even start with this.
Case $n=1$ is obvious,
Assume $n=k$ works, i.e. $(1+x)^k\geq 1+kx$, we want to prove $n=k+1$ also works
$(1+x)^{k+1}=(1+x)(1+x)^k\geq (1+x)(1+kx)$
Since $(1+x)(1+kx)=1+(k+1)x+kx^2\geq 1+(k+1)x$ (because $kx^2$ must be positive), so
$$(1+x)^{k+1}\geq 1+(k+1)x$$
Which is the inductive step we need to prove