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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.

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    @Lazar Goog$l$e Bernoulli's inequality.2012-03-23

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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} $.

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    @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.2012-03-24
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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:

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

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    @BroskianaJones Well, yes. Look at this [question](http://math.stackexchange.com/questions/123655/evaluate-and-prove-by-induction-sum-kn-choose-k-sum-frac1kk1/123665#123665). 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.2012-03-24