Why do we need an integral to prove that $\frac{22}{7} > \pi$? We know this famous (and beautiful) integral which shows that $\dfrac{22}{7} > \pi$ as : 
$$0 < \int_0^1 \frac{x^4(1-x)^4}{1+x^2} \, dx = \frac{22}{7} - \pi$$
Now since the integrand is positive, hence:
$$\dfrac{22}{7}-\pi>0$$
$$\color{blue}{\dfrac{22}{7}>\pi}$$

Although I can see its beauty, why is it needed to show that  $\dfrac{22}{7} > \pi$ ?
Can't we just say that : 
$$\dfrac{22}{7}=\color{red}{3.142}857142857142857\cdots = 3. \overline{142857}$$
$$\pi =\color{red}{3.141}592653589793238\cdots$$
And hence it is greater ??

Thanks!!

If you can derive
$\displaystyle \sum_{n=1}^{\infty}\dfrac{1}{n^2}=\dfrac{\pi^2}{6}$
then you can extract $\pi<(22/7)$ via a comparison test.  It does, however, take a little patience.
First separate the $n=1$ through $n=4$ terms:
$\dfrac{\pi^2}{6}-\dfrac{205}{144}=\displaystyle \sum_{n=5}^{\infty}\dfrac{1}{n^2}$
Then we have
$\dfrac{1}{n^2}<\dfrac{1}{(n-\frac{1}{2})(n+\frac{1}{2})}=\dfrac{1}{(n-\frac{1}{2})}-\dfrac{1}{(n+\frac{1}{2})}$
$\dfrac{\pi^2}{6}-\dfrac{205}{144}<\displaystyle \sum_{n=5}^{\infty}\left(\dfrac{1}{(n-\frac{1}{2})}-\dfrac{1}{(n+\frac{1}{2})}\right)$
The last sum telescopes to $2/9$.  Therefore
$\dfrac{\pi^2}{6}<\dfrac{205}{144}+\dfrac{2}{9}$
$\pi^2<\dfrac{79}{8}$ 
And then
$\left(\dfrac{22}{7}\right)^2=\dfrac{484}{49}=\dfrac{\color{#0055ff}{3872}}{392}$
$\dfrac{79}{8}=\dfrac{\color{#0055ff}{3871}}{392}$
