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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?

shinzou
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  • The first $\to$ should rather be an $\iff$. – Pedro Jan 28 '15 at 20:52
  • @PedroTamaroff why is it wrong to write just $\rightarrow$? isn't it enough for this case? – shinzou Jan 28 '15 at 20:53
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    You want to show that if $0<x<1$, $1-x<\frac{1}{1+x}$. You cannot assume what you want to show is true, show it implies a true fact and conclude that what you assumed is true is... true. Your logic is "Let's show $P$. Assume $P$. Since $P$ implies $Q$ and $Q$ is true, $P$ is true." – Pedro Jan 28 '15 at 20:54
  • Oh yes I see that $\frac{1-x}{1+nx}<\frac{1}{1+nx}$ the problem is with the denominator. – shinzou Jan 28 '15 at 20:57
  • Note that $$\frac{1-x}{1+nx}<\frac{1}{1+(n+1)x}$$ is true if and only if $-(n+1)x^2<0$ (just cross multiply and play around a while). Since $x^2>0$ and $n+1>0$, you're done. – Pedro Jan 28 '15 at 20:57
  • @PedroTamaroff why is it assuming what I need to prove? – shinzou Jan 28 '15 at 21:01

3 Answers3

5

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

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

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

Jack D'Aurizio
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    Ingenious. Similar to the proofs that $(1+1/n)^n$ is increasing and $(1+1/n)^{n+1}$ is decreasing (and so both approach a common limit, which I propose to call "$e$"). – marty cohen Jan 10 '19 at 21:59
  • The proofs are in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563. – marty cohen Jan 10 '19 at 22:03
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Write it in the form $(1+nx)(1-x)^n \lt 1$.

If true for $n$, then

$\begin{array}\\ (1+(n+1)x)(1-x)^{n+1} &=(1+(n+1)x)(1-x)^{n+1}\\ &=\dfrac{(1+(n+1)x)}{1+nx}(1+nx)(1-x)^{n+1}\\ &=\dfrac{(1+(n+1)x)(1-x)}{1+nx}(1+nx)(1-x)^{n}\\ &<\dfrac{(1+(n+1)x)(1-x)}{1+nx}\\ &=\dfrac{1+nx-(n+1)x^2}{1+nx}\\ &\lt 1\\ \end{array} $

marty cohen
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