Proving an inequality by induction and figuring out intermediate inductive steps? I'm working on proving the following statement using induction:
$$ \sum_{r=1}^n \frac{1}{r^2} \le \frac{2n}{n+1} $$
Fair enough. I'll start with the basis step:
Basis Step: (n=1)
$$ \sum_{r=1}^n \frac{1}{r^2} \le \frac{2n}{n+1} $$
$$ \frac{1}{1^2} \le \frac{2}{1+1} $$
$$ 1 \le 1  \checkmark  $$
Inductive Step:
$$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{(n+1)+1} $$
$$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{(n+1)+1} $$
$$ \sum_{r=1}^n \frac{1}{r^2} + \frac{1}{(n+1)^2} \le \frac{2n}{n+1} + \frac{1}{(n+1)^2} $$
$$ \sum_{r=1}^{k+1} \le \frac{2n}{n+1} + \frac{1}{(n+1)^2} $$
My goal is to prove $\forall_{n\ge1} s(n) \implies s(n+1) $ or that this inequality holds true for all $n\ge1$.
I'm not quite sure to go from here on the inductive step. I understand that I need to basically work some clever substitution and manipulation into the problem to end up with:
$$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{(n+1)+1} $$
However, I'm not quite sure what needs to done to obtain this after attempting a few times.

For the inductive step you have to show that $ \sum_{r=1}^{n} \frac{1}{r^2} \le \frac{2n}{n+1} $ implies $ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2n+2}{n+2} $.
Start with $ \sum_{r=1}^{n+1} \frac{1}{r^2}$ and show that is less or equal than $\frac{2n}{n+1}+\frac{1}{(n+1)^2}$ (use the inductive hypothesis). Then, show that the last quantity is less or equal than $\frac{2n+2}{n+2}$. 
