Proving with induction $(1-x)^n<\frac 1 {1+nx}$

Question

Prove using induction that $\forall n\in\mathbb N, \forall x\in \mathbb R: 0<x<1: (1-x)^n<\frac 1 {1+nx}$

My attempt:

Base: for $n=1: 1-x<\frac 1 {1+x}\iff 1-x^2<1$, true since $0<x<1$.

Suppose the statement is true for $n$, prove for $n+1$:

$(1-x)^{n+1}=(1-x)(1-x)^{n}\overset{i.h}<\frac{(1-x)}{1+nx}$

Now I got stuck, maybe another induction to show that $1+nx+x<1+nx$? Is there another way?

Moreover, I was told it's wrong to begin with $(1-x)^{n+1}$ and reach to $\frac 1 {1+(n+1)x}$ but why? Is it assuming what I need to prove?

Answer 1

Apply again the base case: $1-x<\displaystyle\frac1{1+x}$ and that $x^2>0$ to get

$$\frac{1-x}{1+nx} < \frac1{(1+nx)(1+x)}=\frac1{1+(n+1)x+nx^2}<\frac1{1+(n+1)x}\,.$$

Answer 2

Notice that there is no need of induction. By the AM-GM inequality, $$ (1+nx)(1-x)^n < \left(\frac{(1+nx)+n(1-x)}{n+1}\right)^{n+1} =1.$$