I am trying to prove that:
$(1+x)^n \ge 1 + nx$ for every $n \in \mathbb N^+ $ and $\ x \in (-1, \infty)$
I have never seen induction on more that one variable.
Since $(-1, \infty)$ has no least element can I even induct on this? Would strong induction be preferable?
Here is my proof for all $n \in \mathbb N^+$ by inducting on $n$.
Proof:
Suppose $P(n) : (1+x)^n \geq 1 + nx$.
$P(1) = 1+x \ge 1+x$
$P(n) \Rightarrow p(n+1)$
$(1+x)^n \ge (1+nx)$
$(1+x) (1+x)^n \ge (1+nx)(1+x)$
$(1+x)^{n+1} \ge 1+ x + nx + nx^2 $
Since $nx^2 \ge 0 $ for all $n,x$ then:
$1+ x + nx + nx^2 \ge 1 +xn + x$ .
Therefore:
$(1+x)^{n+1}\ge 1 +xn + x$
$(1+x)^{n+1}\ge 1 + x (n+1) $
Therefore:
$(1+x)^n \ge (1+nx) \Rightarrow (1+x)^{n+1} \ge (1+(n+1)x)$
Is this correct? Can anyone provide any guidance on how to approach induction on the interval? Is it even possible?
Cheers Guys!
There is no "double induction". You just do the same induction proof for all $x$. All you are using in your proof is that $x\geq-1$, so any $x\in[-,1\infty)$ would work.