$\sum_{n=1}^{\infty}\frac{1}{n^2} <\frac{33}{20}$ 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. 

Beuker-like integrals are really helpful. We have $\zeta(2)=2\int_{0}^{1}\frac{-\log x}{1+x}\,dx $ and 
$$ \int_{0}^{1}\frac{-\log(x)x^4(1-x)^2}{1+x}\,dx= -\frac{493}{150}+2\,\zeta(2). $$
The LHS is positive and bounded by $\frac{1}{120}$, hence
$$ \zeta(2) \in\left(\frac{493}{300},\frac{659}{400}\right)$$
implies $\zeta(2)=\color{green}{1.64}\ldots$

By creative telescoping™, the inequality $x\leq \text{arctanh}(x)$ for $x\in(0,1)$ and the Weierstrass product for the (hyperbolic)sine function we also have
$$ \zeta(2) \leq \frac{205}{144}+\frac{1}{2}\log\left(\frac{36\sinh\pi}{85\pi}\right) $$
implying $\zeta(2)<1.645=\frac{329}{200}$.
