A simpler proof of $(1-x)^n<\frac{1}{1+nx}$

Question

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.

Answer 1

Hint: for $0<x<1$, the inequality $$ (1-x)^n<\frac{1}{(1+x)^n} $$ is equivalent to $$ (1-x^2)^n<1 $$

Or, mimicking the proof of Bernoulli’s inequality, we have to prove that $$ (1-x)^{-n}>1+nx $$ The statement is true for $n=1$, because it is $1-x^2<1$. Suppose it holds for $n$. Then $$ (1-x)^{-n-1}=(1-x)^{-n}(1-x)^{-1}>\frac{1+nx}{1-x} $$ by the induction hypothesis. We are reduced to prove that $$ \frac{1+nx}{1-x}\ge 1+(n+1)x $$ which becomes $$ 1+nx\ge1+(n+1)x-x-(n+1)x^2 $$

Answer 2

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)

Answer 3

Let $f(x)=(1-x)^{-n}$. Then by applying mean value theorem for $f$ we have,

$$\frac{f(x)-f(0)}{x-0}=f'(\zeta)$$(for some $0<\zeta<x<1$)

$$\frac{(1-x)^{-n}-(1-0)^n}{x-0}=(-n)(1-\zeta)^{-n-1}(-1)=\frac{n}{(1-\zeta)^{n+1}}>n$$

The above line is true as $1-\zeta<1\Rightarrow\frac{1}{1-\zeta}>1$

$$\Rightarrow (1-x)^{-n}-1>nx$$

$$\Rightarrow (1-x)^{-n}>1+nx$$

Now $1-x>0, 1+nx>0$ implies,

$$\Rightarrow \frac{1}{1+nx}>(1-x)^n$$

Answer 4

What about AM-GM?

$$(1-x)^n(1+nx) = \text{GM}(1-x,1-x,\ldots,1-x,1+nx)^{n+1} \color{red}{\leq} \text{AM}(1-x,1-x,\ldots,1-x,1+nx)^{n+1}=1$$ and the inequality holds tight since $(1-x)\neq(1+nx)$. Done.

Answer 5

$(1-x)^n = 1/(1-x)^{-n}=1/(1+nx+n(n-1)x^2/2 +\cdots) < 1/(1+nx)$, assuming $ 0 < x < 1$.