If $0<h<1$ prove that $1/(1+nh)>(1-h)^{n}$

Question

Prove that for $0 < h < 1$ and $n\in\mathbb{N}$, $$\frac{1}{1+nh}>(1-h)^{n}.$$

I am trying with induction , but necessary i would be happy for another way to prove this, here what i got : first we prove that the inequality correct for $n = 1$ when $0 < h < 1$ . Second we assume that the inequality correct for $n$ . then we should prove that the inequality is also correct for $n+1$ .

$$\frac{1}{1+nh+h}>\frac{1-h}{nh+1}\implies \frac{1}{nh+h+1}>\frac{1-h}{nh+1}$$ How to continue from here ? Or there another way to prove this ?

Answer 1

Induction is fine. For the inductive step, note that $$(1-h)^{n+1}=(1-h)^{n}(1-h)<\frac{1}{1+nh}(1-h)$$ where $1-h>0$. So it remains to show that $$\frac{1}{1+nh}(1-h)\leq \frac{1}{1+(n+1)h}$$ that is $$(1+(n+1)h)(1-h)\leq 1+nh$$ or $$1+nh-\underbrace{(n+1)h^2}_{>0}\leq 1+nh$$ which holds.

Answer 2

As an alternative

$$\frac{1}{1+nh}>(1-h)^{n} \iff \frac1{1-h} >(1+nh)^\frac1n$$

and by Bernoulli's inequality

$$(1+nh)^\frac1n <1+h <\frac1{1-h}$$

indeed

$$1+h<\frac1{1-h}\iff 1-h^2<1$$

which is true.

Answer 3

Note that $0\lt h\lt1$ means we can write $h={x\over1+x}$ with $x\gt0$. Doing so turns the inequality to be proved into

$${1\over1+{nx\over1+x}}\gt\left(1\over1+x\right)^n$$

and this is equivalent to

$$(1+x)^{n+1}\gt1+(n+1)x$$

which is easy enough to prove by induction:

$$(1+x)^{1+1}=1+2x+x^2\gt1+2x=1+(1+1)x$$

and $(1+x)^{n+1}\gt1+(n+1)x$implies

$$\begin{align} (1+x)^{(n+1)+1} &=(1+x)(1+x)^{n+1}\\ &\gt(1+x)(1+(n+1)x)\\ &=1+(1+n+1)x+(n+1)x^2\\ &\gt1+((n+1)+1)x \end{align}$$

Remark: One thing to be aware of is that the strict inequality $1/(1+nh)\gt(-1h)^n$ does not hold when $n=0$, so you really need to be a bit careful in saying "$n\in\mathbb{N}$," since some conventions have $0\in\mathbb{N}$.

Answer 4

Another way to prove it is by applying the AM-GM inequality:

$(1-h)^n(1+nh) < \left(\dfrac{(1-h)+(1-h)+\cdots + (1-h) + (1+nh)}{n+1}\right)^{n+1}= \left(\dfrac{n+1}{n+1}\right)^{n+1}=1\implies \dfrac{1}{1+nh}>(1-h)^n$ .