TITLE: Proving $ \left(\frac{n^2+1}{n^2}\right)^n\ge\frac{n+1}{n}$ by induction.
QUESTION [6 upvotes]: 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.

REPLY [4 votes]: 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:
$$\begin{align*}
\left(\frac{1+n^2}{n^2}\right)^n&=\left(1+\frac1{n^2}\right)^n\\
&=\sum_{k=0}^n\binom{n}k\left(\frac1{n^2}\right)^k(1)^{n-k}\\
&=1+n\left(\frac1{n^2}\right)+\dots\\
&=1+\frac1n+\dots\\
&\ge 1+\frac1n\;.
\end{align*}$$