<s>[INST] Inequality related to sum of reciprocals: $\sum_{k=1}^{n} {\frac {1} {k^2}} > \frac {3n}{2n+1}$? 
For every integer $n>1$, prove that :
$\sum_{k=1}^{n} {\frac {1} {k^2}} > \frac {3n}{2n+1}$

I don't seem to find any clue on how to relate the left side of the inequality to the right side.
I tried a little bit of AM-GM on the set {$\frac {1}{1^2}, \frac {1}{2^2},..,\frac {1}{n^2}$} :
$\sum_{k=1}^{n} {\frac {1} {k^2}} \geq n(\frac {1} {{(n!)}^2})^{1/n}$
Is that something helpful?
How to proceed ?
 [/INST] for $n=2$ we have we have $$1+\frac{1}{4}>\frac{6}{5}$$ this is true since we have $$25>24$$
now we assume that is true:
$$\sum_{k=1}^n\frac{1}{k^2}>\frac{3n}{2n+1}$$
and we have to prove that
$$\sum_{k=1}^{n+1}\frac{1}{k^2}>\frac{3(n+1)}{2(n+1)+1}$$
we show that
$$\sum_{k=1}^{n+1}\frac{1}{k^2}>\frac{3n}{2n+1}+\frac{1}{(n+1)^2}>\frac{3(n+1)}{2(n+1)+1}$$
this is true since $$\frac{3n}{2n+1}+\frac{1}{(n+1)^2}-\frac{3(n+1)}{2(n+1)+1}={\frac {n \left( 2+n \right) }{ \left( 2\,n+1 \right)  \left( n+1
 \right) ^{2} \left( 2\,n+3 \right) }}
>0$$ since $n>1$
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