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For any positive integer $n$ and real number $x > -1$, show that $(1 + x)^n > 1 + nx$. This is Bernoulli’s inequality but I can't figure out how to start with this. Can someone help? Thanks

Martin Sleziak
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user142743
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    Like a good neighbor, Wikipedia is there: http://en.wikipedia.org/wiki/Bernoulli%27s_inequality – David H Apr 13 '14 at 02:34
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    The statement as written is false. Take $n = 1$. – Ink Apr 13 '14 at 02:55
  • Actually, all answers got this wrong here. The statement is correct only for $n\ge 2$, $x>-1$ and the inequalities are strict. In contrast, the answers start their induction at $n=0$ and show weak, instead of strict, inequality. – mathse Apr 13 '14 at 03:13
  • See http://math.stackexchange.com/questions/44432/proof-of-bernoullis-inequality or http://math.stackexchange.com/questions/656771/showing-that-1a0-implies-1an-ge-1-na – Martin Sleziak Apr 13 '14 at 04:49

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For $x\boxed{>}-1$ and $n=1$, the statement is $1+x> 1$. Let the I.H. be $(1+x)^n> 1+nx$. For $n+1$, $$(1+x)(1+x)^n>(1+x)(1+nx)=1+(n+1)x+nx^2\geq 1+(n+1)x$$ Q.E.D.

  • Actually, the inequalities are to be strict; $n$ is to be only positive, and the condition $x>-1$ is never used (probably it's true for all $x$ when $n>0$ ...) – mathse Apr 13 '14 at 02:40
  • @mathse Well, since we're talking about Bernoulli's inequality, we need $x\geq-1$. –  Apr 13 '14 at 02:42
  • Still, you shouldn't use $\ge$ when the proof demands $>$, etc. Why do we need $x\ge -1$? What am I missing? – mathse Apr 13 '14 at 02:43
  • What is the first interpretation of $\ge$? The first is strict? – mathse Apr 13 '14 at 02:47
  • In any case, the question demands strict inequality and you show weak inequality. Also you start with $n=0$ (that's why you only get weak inequality) ... – mathse Apr 13 '14 at 02:48
  • $\ge$ does not imply $>$... $5\ge 5$, but $5\not > 5$. – mathse Apr 13 '14 at 02:49
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    @mathse I made a mistake. Please see my edited answer. –  Apr 13 '14 at 02:51
  • Yes, okay, your statements were close to becoming embarrassing ... – mathse Apr 13 '14 at 02:55
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    But, strictly, your proof is still not correct ... For $x>-1$, too, you must exclude the $n=0$ case in order to have strict inequality as the question demands. The only properly correct proof here has been given by @ml0105 ... – mathse Apr 13 '14 at 03:01
  • @mathse I edited my answer. –  Apr 13 '14 at 14:44
  • Come on, @Sanath. You drank too much wine on the weekend? :) Your answer to this extremely simple question is still not correct. – mathse Apr 13 '14 at 15:22
  • @mathse Please tell me where the flaw is. –  Apr 13 '14 at 15:24
  • The relation is correct only for $n\ge 2$. See our comments above. – mathse Apr 13 '14 at 15:25
  • @mathse Ah, yes! I see the flaw. Should I delete my answer? –  Apr 13 '14 at 15:43
  • I would say either this whole thread is closed because it is duplicate, or you simply correct your answer... – mathse Apr 13 '14 at 15:49
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If $n = 1$, then $(1 + x) = 1 + x$. If $n > 2$, use binomial theorem. We get $(1 + x)^{n} = nx + 1 + \sum_{i=0}^{n-2} \binom{n}{i} x^{i}$.

ml0105
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  • What you wrote is true. But to get the claim in the original post, we need to show that $\sum_{i=0}^{n-2} \binom{n}{i} x^{i}>0$, which is not obvious for $x<0$. (Note that the question asks about $x>-1$.) – Martin Sleziak Apr 13 '14 at 04:54
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Let $x>-1$ be arbitrary. Proceed by induction on $n$. The base case holds $n=0$, since $(1+x)^0 = 1 \geq 1 = 1+0 \times x$. Now as our inductive hypothesis assume the statement holds for $n=k$, so that $(1+x)^k \geq 1+kx$. Go from there, if you need help comment.

Mr.Fry
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Possible homework problem, so I'll give a hint: Consider the binomial theorem.

Stella Biderman
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