Just for fun...
Here is a proof that $\displaystyle \frac{22}{7}$ is a better approximation than $\displaystyle 3.14$.
First we consider the amazing and well known integral formula for $\displaystyle \frac{22}{7} -\pi$ (for instance see this page: Proof that 22/7 exceeds pi).
$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx = \frac{22}{7} -\pi$
We will to show that
$0 < \frac{22}{7} -\pi < \pi - 3.14$
That $\displaystyle 0 < \frac{22}{7} - \pi$ follows trivially from the above integral.
We will now show that
$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx < \frac{1}{700}$
We split this up as
$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx = \int_{0}^{\frac{1}{2}}\frac{x^{4}(1-x)^{4}}{1+x^2} + \int_{\frac{1}{2}}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx$
The first integral can be upper-bounded by replacing $\displaystyle x$ in the denominator with $\displaystyle 0$ and the second integral can be upper-bounded by replacing $\displaystyle x$ in the denominator with $\displaystyle \frac{1}{2}$.
Thus we have that
$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx < \int_{0}^{\frac{1}{2}}x^{4}(1-x)^{4}dx + \int_{\frac{1}{2}}^{1} \frac{4x^{4}(1-x)^{4}}{5}dx $
Now $\int_{0}^{\frac{1}{2}}x^{4}(1-x)^{4}dx = \int_{\frac{1}{2}}^{1}x^{4}(1-x)^{4}dx$ as $\displaystyle x^4(1-x)^4$ is symmetric about $\displaystyle x = \frac{1}{2}$
It is also known that $\int_{0}^{1}x^{4}(1-x)^{4}dx = \frac{1}{630}$ (see the above page again)
Thus we have that
$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx < \frac{1}{2*630} + \frac{4}{5*2*630} = \frac{1}{700}$
Thus we have that
$\frac{22}{7} - \pi < \frac{1}{700}$
i.e
2\pi > 2(\frac{22}{7} - \frac{1}{700})
2\pi > \frac{22}{7} + \frac{22}{7} - \frac{2}{700}
2\pi > \frac{22}{7} + \frac{2200}{700} - \frac{2}{700}
2\pi > \frac{22}{7} + \frac{2198}{700}
2\pi > \frac{22}{7} + \frac{314}{100}
Thus we have that
$0 < \frac{22}{7} - \pi < \pi - \frac{314}{100}$