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Show that for $0<a<1$ and $n\in \mathbb N^+$ the following does hold:

$$\left(1-a\right)^{n}<\frac{1}{1+an}$$

I know that for $a \ge -1$ and $n \in \mathbb N^+$ we have that $(1+a)^n \ge 1+an$,on the other hand since $0<a<1$ we have that $0<(1-a^2)^n=(1-a)^n(1+a)^n<1$ which follows that $$(1-a)^n<\frac{1}{(1+a)^n}\le \frac{1}{1+an}$$ $$\implies (1-a)^n< \frac{1}{1+an}$$ Is my solution right?

masaheb
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2 Answers2

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Your proof is nicely done,we could do it like this also :

For $n=1$ its obvious.Suppose its true for $n=k$ then $${(1-a)}^k(1+ak)<1$$ if $n=k+1$ then $$\begin{align*}{(1-a)}^{k+1}(1+ak+a)={(1-a)}^k(1+ak)(1-a)\frac{(1+ak+a)}{1+ak}&\\ <\frac{(1-a)(1+ak+a)}{1+ak}&\\= 1-\frac{a^2k+a^2}{1+ak}<1 \end{align*}$$

Hence its true for $n=k+1$ our proof by induction is over.

Albus Dumbledore
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Let $$f(a)=(1-a)^n(1+an),a\in[0,a].$$ Compute its derivative: $$f'(a)=-n(n+1)a(1-a)^{n-1}<0,\forall a\in(0,1).$$ So $$0=f(1)<f(a)=(1-a)^n(1+an)<f(0)=1,\forall a\in(0,1),$$ which implies $$\left(1-a\right)^{n}<\frac{1}{1+an}.$$

Riemann
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