TITLE: Proving the following inequality using induction?
QUESTION [6 upvotes]: I need to prove by induction that $Q_n \ge\frac{10}{3}-\frac5{3n}$ for $n\ge2$$$
Q_n = \left(\frac21\right)\cdot\left(\frac54\right)\cdot\left(\frac{10}{9}\right)\cdot\left(\frac{17}{16}\right)\cdot\ldots\cdot\left(\frac{n^2+1}{n^2}\right)=\prod^n_{k=1}\left(\frac{k^2+1}{k^2}\right)$$
I have been trying to come up with a formula for $Q_n$ , I think I need to know that before I can go ahead with the proof right?  I did try writing $Q_n$ as: $$Q_n = \left(1+\frac{1}{1^2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{3^2}\right)\ldots\left(1+\frac{1}{n^2}\right)$$But that's not really helping me come up with a formula or the proof.  Any help would be great.

REPLY [3 votes]: The point of induction is that you don't need to find a nicer formula for $Q_n$. You prove the base case for $n=2$ by plugging it in and verifying it. You then assume it's true for $Q_n$, and then you just need to show that $Q_{n+1}\geq \frac{10}{3}-\frac{5}{3(n+1)}$ follows from $Q_n\geq \frac{10}{3}-\frac{5}{3n}$. It does help to see that $Q_{n+1}=\left(1+\frac{1}{(n+1)^2}\right)Q_n$. Plug in the bound you have for $Q_n$ and then simplify. It amounts to showing that
$$\left(1+\frac{1}{(n+1)^2}\right)\left(\frac{10}{3}-\frac{5}{3n}\right)\geq\frac{10}{3}-\frac{5}{3(n+1)}$$