<s>[INST] A simpler proof of $(1-x)^n<\frac{1}{1+nx}$ I proved the following inequality: 
Let $x\in\mathbb{R}, 0<x<1, n\in\mathbb{N}$, then $(1-x)^n<\frac{1}{1+nx}$ 
however, judging from the context in the exercise book, I feel like there is a much simpler way to prove it, but I can't see it. So I'm asking for that simpler alternative proof, or at least a hint to it. My proof as follows:
Bernoulli inequality states that for $-1<x, x\neq 0, n\in \mathbb{N},n>1$ the following is true:$(1+x)^n>1+nx$.
Thus, $\frac{1}{(1+x)^n}<\frac{1}{1+nx}$ is also true. 
Then I need to show that $(1-x)^n<\frac{1}{(1+x)^n}$, which is equivalent to $\frac{1}{(1-x)^n}>(1+x)^n$, which I prove by induction:
Basecase: $n=1$ $1=\frac{1-x}{1-x}\Leftrightarrow 1=\frac{1}{1-x}-\frac{x}{1-x}\Leftrightarrow 1+\frac{x}{1-x}=\frac{1}{1-x}$
Let $a,b\in \mathbb{R_{>0}}$ and $0>b>1$ Then $\left[ a>ab \right]\Leftrightarrow \left[a<\frac{a}{b}\right]$. Thus $\left[ 0<x<1\right] \Rightarrow \left[ x<\frac{x}{1-x}\right]$
Thus $1+\frac{x}{1-x}=\frac{1}{1-x} \Rightarrow 1+x<\frac{1}{1-x} \square$
Inductive step: Assume $(1+x)^n<\frac{1}{(1-x)^n}$. Need to show $(1+x)^{n+1}<\frac{1}{(1-x)^{n+1}}$. 
$(1+x)^{n+1}<\frac{1}{(1-x)^{n+1}}\Leftrightarrow (1+x)^n\cdot(1+x)<\frac{1}{(1-x)^n} \cdot \frac{1}{1-x}$
Let $a,b,c,d \in \mathbb{R_{>0}}$. Then $[a>c]\wedge[b>d] \Rightarrow [ab>cd]$. $(1+x)^n<\frac{1}{(1-x)^n}$ was the assumption and $1+x>\frac{1}{1-x}$ was the basecase, therefore $(1+x)^n\cdot(1+x)<\frac{1}{(1-x)^n} \cdot \frac{1}{1-x} \Leftrightarrow (1+x)^{n+1}<\frac{1}{(1-x)^{n+1}} \square$
Thus $\frac{1}{(1+x)^n}<\frac{1}{1+nx}$ and $(1-x)^n<\frac{1}{(1+x)^n}$ are both true, which implies the original statement $(1-x)^n<\frac{1}{1+nx} \square$ If I were to count the proof of the Bernoulli inequality by induction, it would mean that I used induction twice in order to prove something that basic, which to me doesn't seem to be a sensible thing to do.
 [/INST] First note: 
$$\frac{1+nx}{1-x}=\frac{(1+nx)(1+x)}{(1-x)(1+x)}=\frac{1+(n+1)x+nx^2}{1-x^2}\ge\frac{1+(n+1)x}{1}$$
So: 
$$\frac{1+nx}{1+(n+1)x}\ge1-x$$
Take a product of these expressions for $n=0,1,\cdots,N-1$ to see
$$\frac{1}{1+Nx}\ge(1-x)^N$$
(this can easily be recast as an induction)
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