Simplest way to get the lower bound $\pi > 3.14$ Inspired from this answer and my comment to it, I seek alternative ways to establish $\pi>3.14$. The goal is to achieve simpler/easy to understand approaches as well as to minimize the calculations involved. The method in my comment is based on Ramanujan's series $$\frac{4}{\pi}=\frac{1123}{882}-\frac{22583}{882^{3}}\cdot\frac{1}{2}\cdot\frac{1\cdot 3}{4^{2}}+\frac{44043}{882^{5}}\cdot\frac{1\cdot 3}{2\cdot 4}\cdot\frac{1\cdot 3\cdot 5\cdot 7}{4^{2}\cdot 8^{2}}-\dots\tag{1}$$ This is quite hard to understand (at least in my opinion, see the blog posts to convince yourself) but achieves the goal of minimal calculations with evaluation of just the first term being necessary.
On the other end of spectrum is the reasonably easy to understand series
$$\frac\pi4=1-\frac13+\frac15-\cdots\tag2$$
But this requires a large number of terms to get any reasonable accuracy. I would like a happy compromise between the two and approaches based on other ideas apart from series are also welcome.

A previous question of mine gives an approach to estimate the error in truncating the Leibniz series $(2)$ and it gives bounds for $\pi$ with very little amount of calculation. However it requires the use of continued fractions and proving the desired continued fraction does require some effort.

Another set of approximations to $\pi$ from below are obtained using Ramanujan's class invariant $g_n$ namely $$\pi\approx\frac{24}{\sqrt{n}}\log(2^{1/4}g_n)\tag{3}$$ and $n=10$ gives the approximation $\pi\approx 3.14122$ but this approach has a story similar to that of equation $(1)$.

Approach-1: Using basic geometry (This method is inspired by Archimedes method to estimate $\pi$)
Draw a circle of radius 'r' . 
Now, draw a regular polygon (say, polygon-1) with '$N_c$' number of sides such that the circle is circuim-cicle to this polygon. 
Also, draw another regular polygon (say, polygon-2) with '$N_i$'number of sides such that the circle is in-circle to this polygon.
So,perimeter of Polygon-1 is equal to $C_1=2rN_c sin(\frac{\pi}{N_c})$ 
Also, it can be found that perimeter of Polygon-2 is equal to $C_2=2rN_i tan(\frac{\pi}{N_i})$ 
If 'C' is the circumference of the circle, then we know that, $C=2r\pi $
So, by using the fact that 'the sum of any two two sides of a triangle is always grater than the third side' , we can easily show that $C_1<C<C_2$ 
Putting values, we find that, $2rN_c sin(\frac{\pi}{N_c})<2r\pi<2rN_i tan(\frac{\pi}{N_i})$ 
This simplifies as, $N_c sin(\frac{\pi}{N_c})<\pi<N_i tan(\frac{\pi}{N_i})$ .......(1)
So, for any two assumed positive values of $N_c$ and $N_i$ , the relation (1) is always valid and is satisfied.
As we can observe easily, if $N_c$ and $N_i$ gets infinitely larger, then both polygons-1 & 2 as mentioned above approximates the circle of radius 'r'more closely.
Let, $N_c=N_i=N$ , then it can be proved very easily that, $ \lim_{{N}\to{\infty}} Nsin(\frac{\pi}{N})=\pi$ 
 and $ \lim_{{N}\to{\infty}} Ntan(\frac{\pi}{N})=\pi$ 
So, inequality (1) is the simplest way to get lower and upper bounds for $\pi$ and these bounds can be improved as per our choice of values of$N_c$ and $N_i$ as per above discussion.
For example. $N_c=N_i=N=4$ gives a bound as, $4 sin(\frac{\pi}{4})<\pi<4 tan(\frac{\pi}{4})$ i.e. $2\sqrt{2} <\pi<4 $
Also, if we increase $N$ to $N_c=N_i=N=6$ , we get a bound as, $6 sin(\frac{\pi}{6})<\pi<6 tan(\frac{\pi}{6})$ i.e. $3 <\pi<2\sqrt{3} $ which is a more tighter bound and so on.
It can be found numerically that, when N>57, we have $3.14 <\pi $
Also, if $N_c=6 , N_i=4$ then, using the inequality $N_c sin(\frac{\pi}{N_c})<\pi<N_i tan(\frac{\pi}{N_i})$ , we find that $6 sin(\frac{\pi}{6})<\pi<4 tan(\frac{\pi}{4})$ i.e.$ 3<\pi<4 $ 
So, $\pi$ can be approximated as the average of these limits as $N_c,N_i\to\infty$ 
Approach-2: Riemann Zeta function:
We know a well known and famous relation that was given by Euler that,
if 'n' is a positive integer and if $\zeta(2n)=\frac{1}{1^{2n}}+\frac{1}{2^{2n}}+\frac{1}{3^{2n}}+\frac{1}{4^{2n}}+.....$  then, $\zeta(2n)=(-1)^{n+1}\frac{B_{2n}2^{2n-1}\pi^{2n}}{(2n)!}$ 
Separating $\pi$ from this relation, we get $\pi^{2n}=\frac{(2n)!\zeta(2n)}{(-1)^{n+1}B_{2n}2^{2n-1}}$ 
So, $\pi=(\frac{(2n)!\zeta(2n)}{(-1)^{n+1}B_{2n}2^{2n-1}})^{\frac{1}{2n}}$ 
As it can be observed easily that $\lim_{{n}\to{\infty}}\zeta(2n)=1$
Hence it can be concluded that
$\pi=\lim_{{n}\to{\infty}}(\frac{(2n)!}{(-1)^{n+1}B_{2n}2^{2n-1}})^{\frac{1}{2n}}$ 
Since $\sum_{r=1}^{k}\frac{1}{r^{2n}}<\zeta(2n)$ , hence 
$\pi>(\frac{(2n)!\sum_{r=1}^{k}\frac{1}{r^{2n}}}{(-1)^{n+1}B_{2n}2^{2n-1}})^{\frac{1}{2n}}$
Actual application of this relationship is quite computationally time consuming but it's really a good approximation when 'n'and 'k' are fairly high values (even $n>8$ and $k>10 $ will give a reasonably good and closer higher bound for $\pi$) 
