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Prove for $n\in\mathbb{N}$: $$ \left(\frac{n^2+1}{n^2}\right)^n\ge\frac{n+1}{n}.$$ by induction.

I'm doing induction ahead of my regular classes because I need it for competition coming in few months. I've been introduced to induction before, but I've never proved inequalities with it before, so I'm pretty new to this, especially since I have $n$ in both power and base.

Lazar Ljubenović
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2 Answers2

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If you particularly want to prove it by induction, that’s certainly possible, as Broskiana Jones has just demonstrated. However, the binomial theorem will do the trick for you without induction:

$$\begin{align*} \left(\frac{1+n^2}{n^2}\right)^n&=\left(1+\frac1{n^2}\right)^n\\ &=\sum_{k=0}^n\binom{n}k\left(\frac1{n^2}\right)^k(1)^{n-k}\\ &=1+n\left(\frac1{n^2}\right)+\dots\\ &=1+\frac1n+\dots\\ &\ge 1+\frac1n\;. \end{align*}$$

Brian M. Scott
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    @quartz: Broskiana could have stated it more clearly, but since the result is proved simultaneously for all $a>-1$, the fact that $a$ is a function of $n$ in the original problem doesn’t matter; see Ilya’s response to you under the other answer. – Brian M. Scott Mar 23 '12 at 19:08
  • I like this answer because straightforward proofs rather than inductive ones tend to be much more appreciable and elegant to me. – 000 Mar 24 '12 at 04:38
  • @user22144 But then, it is easier to prove the principle of induction than the Binomial Theorem. It depends on the ammo you want to use. – Pedro Mar 24 '12 at 04:58
  • @PeterT.off, very true. I find that the binomial theorem is rather 'normal' and 'appropriate' to me, so you've got it. It's all about the ammo. – 000 Mar 24 '12 at 05:54
  • i would prefer this direct approach also, if the question doesn't require using induction ;D – Geralt of Rivia Mar 24 '12 at 06:25
  • @BroskianaJones Well, yes. Look at this question. Some users propose the use of differential calculus, specially, differentiating a sum and getting a value. I proposed the use of a simple identity ${n \choose k}={n \choose n-k}$. Someone else took an even easier path. I believe sometimes it is better to avoid the use of "higher" tools in easy problems because it makes them look bigger than they are. – Pedro Mar 24 '12 at 06:33
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How about this: We prove something stronger with induction on n :
$ (1 + a)^n \geq 1+ na $ (1) for every $n \in \mathbf{N} $ and a is any fixed real number that is not less than -1.
It's true for n = 1
Suppose $ (1+a)^k\geq1+ka $
then $(1+a)^{k+1}\geq (1+ka)(1+a) $ since $ 1+a \geq 0$ which means $(1+a)^{k+1}\geq 1+ka + a + ka^2 \geq 1+(k+1)a $.
Therefore (1) is true. Every number $a = 1/n^2 $ satisfies (1)'s condition, so we have
$(1+\frac{1}{n^2})^n\geq1+\frac{1}{n}$ for every $n \in \mathbf{N} $.

Geralt of Rivia
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  • @quartz: the answer is write for any $a$ - that means that is also be given in the form $(1+a_n)^n\geq 1+na_n$ – SBF Mar 23 '12 at 19:05
  • the dependence of a on n doesn't change the fact that (1) is true for $a \geq -1$, particularly it is true for every number a of the form $1/n^2$ where n is the power itself. – Geralt of Rivia Mar 23 '12 at 19:05
  • then it is right. thanks IIya and Broskiana. – quartz Mar 23 '12 at 19:07
  • @Ilya, still I believe that you cannot write induction step $(1+a_1)^{k+1}\geq (1+ka_2)(1+a)$, because $a_1$ and $a_2$ are different. I think answer require one more step that $1+ka_2 \geq 1+ka_1$. – quartz Mar 23 '12 at 19:11
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    @quartz: it was already proved by induction for a fixed $a$. When we apply it to the problem in OP we can change $a$ because we don't use an induction anymore – SBF Mar 23 '12 at 19:13
  • @quartz: i stated that it was an induction on n, and it is true for any $a \geq -1 $. Yeah, perhaps i can make it clearer. – Geralt of Rivia Mar 24 '12 at 04:10