$\sum_{n=1}^{\infty}\frac{1}{n^2} <\frac{33}{20}$ using elementary inequalities There are many ingenious ways for proving $$\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6} \approx 1.6449$$
Using the inequality $$\frac{1}{n^2} < \frac{1}{n^{2}-\frac{1}{4}} =\frac{1}{n-\frac{1}{2}}-\frac{1}{n+\frac{1}{2}}$$ we can see that $$\sum_{n=2}^{\infty}\frac{1}{n^2} <\frac{2}{3} \Rightarrow \zeta(2)<\frac{5}{3}=1.6666$$
Can we improve upon these bounds using elementary inequalities? Like is it possible to show (of course without assuming $\zeta(2)\approx 1.64449$) that $\zeta(2)<\frac{33}{20}$?
If there are much nicer bounds which follow using elementary inequalities I would be happy to see them. 

You only need a few more terms:
$$\zeta(2)<1+\frac14+\frac19+\sum_{n=4}^\infty\frac1{n^2-1/4}
=1+\frac14+\frac19+\frac27$$
which is already less than $33/20$.
