Using the proposition I have checked it for a($1$) $1+x=1+x$ so a($1$) is true A($n$)=$(1+x)^n\ge 1+nx$ assuming this is right according to inductive hypothesis Considering a($n+1$) $( 1+x)^n+1\ge 1+(n+1)x$....($1$) Now $(1+x)^n+1\ge (1+nx)(x+1)$... $(1+x)^{n+1}\ge 1+x(n+1)+nx^2$....($2$) Considering the equation $1$ and $2$ $1+(n+1)x\ge 1+x(n+1)+nx^2\le (1+x)^n+1$ as $nx^2$ is a positive quantity..so the proposition is true...but is the proposition is true if we take $n<0$?
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Have you heard of Bernoulli's inequalities? – Simply Beautiful Art Sep 08 '17 at 15:08
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I know this is Bernoulli's inequality...but I am confused – Akul sharma Sep 08 '17 at 15:10
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4Please use MathJax. – José Carlos Santos Sep 08 '17 at 15:13
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we have to prove that $$ (1+x)^{n+1}\geq 1+(n+1)x$$ multiplying $$(1+x)^n\geq 1+nx$$ by $1+x>0$ we get $$(1+x)^{n+1}\geq (1+nx)(1+x)=1+x(n+1)+nx^2$$ and this is greater than $$1+(n+1)x$$
Dr. Sonnhard Graubner
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Does your post answer the question:"$\texttt{...but is the proposition is true if we take $n<0$? }$" ? – callculus42 Sep 08 '17 at 15:23
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@Akulsharma Is your question answered ? If yes, you can check the answer. – callculus42 Sep 08 '17 at 15:37
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@callculus...the answer given here I already know...but I am expecting different answer...if the proposition holds true for n<0 – Akul sharma Sep 08 '17 at 15:40
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@Akulsharma So my intuition was right. I´m wondering about people who vote up without reading/understanding the question and the answer. – callculus42 Sep 08 '17 at 15:49
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see here for the other cases https://en.wikipedia.org/wiki/Bernoulli%27s_inequality – Dr. Sonnhard Graubner Sep 08 '17 at 16:03
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