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 ?
 A: A cheeky solution:
$$\frac{3n}{2n+1}<\frac{3n}{2n}=\frac32$$
Likewise, for $n\ge7$,
$$\sum_{k=1}^n\frac1{k^2}\ge\sum_{k=1}^7\frac1{k^2}>\frac32>\frac{3n}{2n+1}$$
and one can numerically show this is true for $1<n<7$.
A: Induction, with the step being:
$$\sum_{k=1}^n\frac1{k^2}=\sum_{k=1}^{n-1}\frac1{k^2}+\frac1{n^2}\stackrel{Ind. \;Hyp.}>\frac{3(n-1)}{2(n-1)+1}+\frac1{n^2}=\frac{3n-3}{2n}+\frac1{n^2}=$$
$$=\frac{3n^2-3n+2}{2n^2}\ge\frac{3n}{2n+1}\iff6n^3-4n^2+n+2\ge6n^3$$
and the last inequality must be clear since
$$4n^2-n-2\ge0\;\;\; \forall\, n\ge\frac{1+\sqrt33}8\approx\frac{1+5.6}8\;,\;\;\text{and}\;\;0<\frac{1+5.6}8<1$$
so the inequality is true for all $\;n\in\Bbb N\;$ .
A: 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 $n>1$
