prove, using induction that for natural $n$ and $0<x<1$ that $(1-x)^n<\frac{1}{1+nx}$

Question

How to prove, using induction, that for every natural $n$, and for every $0<x<1$ :$$(1-x)^n<\frac{1}{1+nx}$$

Answer 1

It is just a consequence of the AM-GM inequality, since $x\in (0,1)$ grants:

$$ (1-x)^n (1+nx) \color{red}{\leq} \left(\frac{n\cdot(1-x)+(1+nx)}{n+1}\right)^{n+1}=1 $$ but equality is achieved only when $(1-x)=(1+nx)$, i.e. at $x=0$.

Answer 2

It is not really hard: suppose by the inductive hypothesis that, for some $n$, we have $\;(1-x)^n<\dfrac{1}{1+nx}$. As $0<1-x<1$, multiplying both sides by $1-x$, we get $$(1-x)^{n+1}<\dfrac{1-x}{1+nx}$$ so it is enough to prove $\;\dfrac{1-x}{1+nx}\le\dfrac1{1+(n+1)x}$. As all numbers are positive, this amounts to proving $$(1-x)\bigl(1+(n+1)x\bigr)\le 1+nx\iff 1+nx-(n+1)x^2\le1+nx. $$