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Use mathematical induction to prove the following statement:

For all $b\in\mathbb R$, and for all $n\in\mathbb N$, $$b>-1\implies (1+b)^n \geq 1+nb$$

When $n=1$, the inequality still holds $1+b \geq 1+b$.

For n+1$: $$(1+b)^{n+1} \geq 1+(n+1)b$$ Here I'm not sure the best way to simplify... $$(1+b)^n(1+b)\geq 1+bn+b$$

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    $\begingroup$ Hi and welcome to the site! Since this is a site that encourages learning, you will get much more help if you show us what you have already done. Could you edit your question with your thoughts and ideas? $\endgroup$ – 5xum Sep 25 '14 at 13:01
  • $\begingroup$ How do you format questions? People fix mine but I'd like to post them right in the first place.. $\endgroup$ – user178359 Sep 25 '14 at 13:15
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    $\begingroup$ See this site for hints about nice formatting: meta.math.stackexchange.com/questions/5020/… $\endgroup$ – 5xum Sep 25 '14 at 13:16
  • $\begingroup$ For help on notation, see math.stackexchange.com/help/notation $\endgroup$ – James Holbert Sep 25 '14 at 13:21
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$$(1+b)^{n+1} =(1+b)^n (1+b) \geq (1+nb)(1+b) =1+(n+1)b +nb^2 \geq 1+ (n+1)b.$$

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A few points:

  • The base case, in this case, is $n=0,$ so that's what you should verify, not $n=1.$

  • $$\color{green}{(1+b)^{n+1}} \equiv \underbrace{(1+b)^n(1+b) \geq (1+bn)(1+b)}_{\text{induction hypothesis}} \equiv 1+(n+1)b+\underbrace{b^2n}_{\geq 0} \color{green}{\geq 1+(n+1)b} .$$

  • Never say something like "let $n=n+1$". It makes no sense, algebraically. You can say that if the result is true for $n$, then it's true for $n+1$, or you can say "if the result is true for $n=k$, then it's true for $n=k+1.$"

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  • $\begingroup$ How is the base case n=0? I'm confused I thought 0 was not a natural number. $\endgroup$ – user178359 Sep 25 '14 at 21:15
  • $\begingroup$ You want to prove it for all integers $n>-1.$ The smallest integer greater than $-1$ is $0.$ $\endgroup$ – beep-boop Sep 25 '14 at 21:33
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For the induction step, if $(1+b)^n\geq 1+bn$, then $$ (1+b)^{n+1}-(1+b(n+1))\geq (1+b)(1+bn)-(1+b(n+1))=b^2n\geq 0. $$ Note that the first inequality above uses both the induction hypothesis and $b>-1$.

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