<s>[INST] Proving an inequality by induction, how to figure 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:
$$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{(n+1)+1} $$
$$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{n+2} $$
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:
$$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2n}{n+1} $$
However, I'm not quite sure what needs to done to obtain this after attempting a few times.
 [/INST] I is true that $\displaystyle\sum_{r=1}^n \frac{1}{r^2} \leq \frac{2n}{n+1} $, so we need to prove that $\displaystyle \sum_{r=1}^{n+1} \frac{1}{r^2} \leq \frac{2(n+1)}{n+2}$. It is indeed true: $\displaystyle \sum_{r=1}^{n+1} \frac{1}{r^2} =\displaystyle\sum_{r=1}^n \frac{1}{r^2}+\frac{1}{(n+1)^2}\leq \frac{2n}{n+1}+\frac{1}{(n+1)^2}=\frac{2n(n+1)+1}{(n+1)^2}\leq \frac{2(n+1)}{n+2}$ since $2(n^3+3n^2)+5n+2=\bigg(2n(n+1)+1\bigg)(n+2)\leq 2(n+1)^3=2(n^3+3n^2)+6n+2$.
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