<s>[INST] Prove that $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} > \frac{3n}{2n+1}$ for all $n \geq 2$ by induction 
Prove that $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} > \frac{3n}{2n+1}$ for all $n \geq 2$ by induction

I tried to add $\frac{1}{(k+1)^2}$ to both sides of this inequality assuming it's true for $n = k$, but eventually it resulted in a complex expression in the second side:
$$
\frac{3k}{2k+1} + \frac{1}{(k+1)^2} = \frac{3k^3+6k^2+5k+1}{(k+1)^2 (2k+1)}
$$
that I'm not sure how to finish the proof using it.
Am I going the right way? If so, how can I finish this proof?
and is there an easier way to prove this by induction without facing such confusing expressions?
Thanks.
 [/INST] Now, $$\frac{3k}{2k+1}+\frac{1}{(k+1)^2}-\frac{3(k+1)}{2k+3}=\frac{k(k+2)}{(k+1)^2(2k+1)(2k+3)}>0$$ and by induction we are done!
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