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I am familiar with Bernoulli's inequality which is quite straightforward to prove using induction, but this problem (although simpler at first glance) seems to be more complicated. What do you guys think?

Please stop posting solutions on proving Bernoulli's inequality. I am interested in using only induction to solve the stated problem.

Can anyone solve this problem directly using only induction?

Yes I am aware the inequality is actually strict, but I have just copied the problem directly from a high school textbook.

I am not interested in just a solution, but more interested in how would a high school student who just learned about induction solve this problem?

Matt
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  • Have you tried binomial expansion$?$ – LOL Jul 06 '22 at 16:34
  • can you pls provide the pair $(a,n)$ when the equality holds$?$ – LOL Jul 06 '22 at 16:41
  • Sometimes it's easier to prove by induction something stronger. – J.G. Jul 06 '22 at 16:43
  • A pair where the inequality holds would be $(a,n)=(1,1)$. – Matt Jul 06 '22 at 16:46
  • @Matt are you sure? – LOL Jul 06 '22 at 16:52
  • @LOL yes I believe $(1+1)^1=2 \ge 1=1(1)$ – Matt Jul 06 '22 at 16:55
  • @N.F. Taussig No it does not. As I said, I am familiar with Bernoulli's inequality and how to prove it using induction. The problem I have stated (although seemingly simpler at first glance) appears to be more complex. I am interested if anyone can solve this problem directly using induction. – Matt Jul 06 '22 at 16:59
  • @Matt i asked when will $(1+a)^n=na$ – LOL Jul 06 '22 at 17:01
  • @LOL sorry I misread your comment. For $n \in \mathbb{N}$ and $a>0$ there does not exist a pair $(a,n)$ such that $(1+a)^n=na$. Thus, we could state the problem with a strict inequality, but the problem I came across was stated with $\ge$ instead. – Matt Jul 06 '22 at 17:05
  • @Matt would you be satisfied with the argument that (prove $(1+x)^n\geq 1+nx$)$\land(1+nx\gt nx)$? – insipidintegrator Jul 06 '22 at 17:09
  • @insipidintegrator I am interested if the inequality can be proved directly using induction. – Matt Jul 06 '22 at 17:14

3 Answers3

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First of all I would like to tell that the equality will never hold true. So the real question is to prove $$(1+a)^n>na$$ by induction. Now coming to the question, we can see that at $n=1$ the inequality is true. Assume that, at $n=n$ the inequality is true. Now let's see at $n=n+1$

RHS$=na+a$ it will be maximum when both $na$ and $a$ are maximum. Since, $a$ is a constant it's value depends entirely on $na$. This will be maximum when it is approaching to $(1+a)^n$ For a time being let's assume it's value as $(1+a)^n$

So, $RHS_{max}=(1+a)^n+a$

LHS$=(1+a)^{n+1}=(1+a)^n(1+a)$

Let $(1+a)^n=k$

So, $RHS_{max}=k+a$ and LHS$=(1+a)^{n+1}=k(1+a)=k+ka$

RHS=LHS iff $ka=a$ which will only happen if $k=1$ or $a=0$ But $a$ cannot be $0$ as it is given that $a>0$ So $k$ has to be $1$ which means that $(1+a)^n$ has to be $1$ which can happen if $n=0$ or $a=0$.

Again $n \in \mathbb{N}$ and $a>0$ therefore it can't happen.

$\implies$ $RHS_{max}<LHS$ or $(1+a)^n+a<(1+a)^n(1+a)$ or $$(n+1)a<(1+a)^{n+1}$$

So, we see that even if RHS is equal to $(1+a)^n+a$ it is less than LHS. Since actually RHS $<(1+a)^n+a$ so it means that

$$RHS <(1+a)^n+a<(1+a)^{n+1}=LHS$$

LOL
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Edit:This is the proof of Bernoulli's inequality. I don’t really answered the real question.

Here's my approach. Let’s prove a stronger proposition, namely, $(1+a)^n\ge 1+na$ for $n\in\mathbb{N}$ and $a>0.$ For the base case $n=1$, the proof is trivial. Suppose, that $\forall a>0((1+a)^k\ge 1+ka)$ for some integer $k>1$. Now we claim that $\forall a>0((1+a)^{k+1}\ge 1+(k+1)a)$. Since $(1+a)^{k+1}=(1+a)^k(1+a)\ge (1+ka)(1+a)\ge1+(k+1)a$ for all $a>0$, it follow that our claim is true, this closed the induction.

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    Uh..$(ka)(a)=(k+1)a$…you might want to review that. – insipidintegrator Jul 06 '22 at 17:10
  • Right... I will edit it. –  Jul 06 '22 at 17:12
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    $(ka)(a)\neq (k+1)a$ – Matt Jul 06 '22 at 17:12
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    I’m rolling it back to the previous edit. Your English is absolutely fine IMO. – insipidintegrator Jul 06 '22 at 17:12
  • I have edited the proof correctly. Thanks. –  Jul 06 '22 at 17:23
  • As I have already stated, I am aware of Bernoulli's inequality (the inequality you have proved). I am interested if any can prove the stated inequality directly using only induction. – Matt Jul 06 '22 at 17:26
  • Sorry I don’t read your question in detail, should I delete my answer? –  Jul 06 '22 at 17:30
  • No that is fine. It is still a solution, but I was more interested in solving the problem directly. – Matt Jul 06 '22 at 17:42
  • I have another idea. In the first version of the answer I have claimed incorrectly that $(1+a)^{k+1}=(1+a)^k(1+a)\ge (ka)(a)\ge (k+1)a$, but notice that $(1+a)^{k+1}=(1+a)^k(1+a)\ge (ka)(1+a)\ge (k+1)a$ is true if $ka\ge 1$. And for $ka<1$ we have $(1+a)^{k+1}>(1+a)>ka+a=(k+1)a$. –  Jul 06 '22 at 18:30
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Here is my solution.

First note for $n=1$ we have $$(1+a)^1=1+a \ge a=(1)a$$

We now assume $\exists k \in \mathbb{N}$ such that $$(1+a)^k\ge ka \hspace{10mm}(*)$$

Finally, consider $$\begin{eqnarray}(1+a)^{k+1} &=&(1+a)(1+a)^k \\ &=&(1+a)^k+a(1+a)^k \hspace{10mm} \text{Distribute $(1+a)^k$}\\ &\geq& (1+a)^k+a \hspace{27mm} \text{Since } (1+a)^k>1 \\ &\ge & ka +a \hspace{38.4mm}\text{By }(*) \\ &=&(k+1)a \end{eqnarray}$$

Hence, by the principle of mathematical induction we can conclude $(1+a)^n\ge na$ for all $n\in \mathbb{N}$.

Matt
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