Prove by induction that $\forall x \in \mathbb{R}, x \geq -1, \forall n \in \mathbb{N},n \geq 0$ that $$(1+x)^n \geq 1+nx$$ First of all I have a problem with x being a real number, how can I use induction over a real number? Can I simply assume that it holds for all $y \in [-1,x]$ and then show that it also holds for $x + \epsilon, \epsilon \in \mathbb{R},\epsilon \gt 0$?
If that were possible I would try to prove over n. Prove the base case for n, use induction over x and then prove the induction step over n.
But I have some problems with the induction step over n.
Assuming that it works for all $l \in \mathbb{N}, l \leq n-1$. I need to show $$(1+x)^n=(1+x)^{n-1}(1+x) \geq 1+(n-1)x+x$$ which basically results in showing that $$z(1+x) \geq z + x$$ for some quantity z, correct?
But then I'd get $$zx \geq x$$, which would obviously depend on the value of z.
I'm kind of confused at this point, any help would be appreciated.
Once you get to $$(1+x)^n=(1+x)^{n-1}(1+x)$$ you have rearranged the expression in a form in which you can use the inductive hypothesis, because you can assume that $$(1+x)^{n-1}\ge 1+(n-1)x$$
You should substitute that information into the expression you have for $(1+x)^n$ and see where it gets you when you follow it through logically.
So in comments you suggest the base case $n=0$ which is a good start, because you are asked to prove the statement for $n\ge 0$ - so you have to start at the bottom, or deal with special cases, so let's start at the bottom.
You have $$(1+x)^0=1\ge 1+0\cdot x$$You suggest in your comment that the base case does not work for $x\lt 0$, which is not quite correct - the situation is better than that, because you can raise any positive real number to the power zero. So the condition is $(1+x)\ge 0$ which is equivalent to $x\ge -1$, and you are on your way.