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 $(1)$.

Do you want a solution that is a compromise between calculation efficiency and ease of understanding?  This story might help.
Many, many years ago in school, I was introduced to Fortran and had limited access to a computer.  It was physically large but very low power by today's standards.  I knew the famous $\frac{\pi}{4} = \tan^{-1}(1)$ formula but I also realised how slowly it converges.  I would not be able to get far with the run time available to me.  
The internet did not exist yet and the school's and the local public library did not help.  I knew how the $\frac{\pi}{4}$ was derived so I played with other trig formulae.  I managed to calculate the Taylor series of $\sin^{-1}$ by a mixture of messy differentiation and induction.  I figured that if I evaluated  $\frac{\pi}{6} = \sin^{-1}(\frac{1}{2})$, the convergence would be linear: twice the terms would give me twice the number of decimal places.  In practice, the algorithm was quadratic since if I aimed at $n$ times as many decimal places, I would need $n$ times as many terms and each calculation would take $n$ as long.  
I forget the run time that I had but the best I achieved was 500 decimal places.  
Many years later but still long ago, I rewrote the program in C and ran it on an idle Unix system at work.  In a month, it calculated a million decimal places. For comparison, I ran it on my laptop a few years ago, a million places took 3.5 hours.  My Raspberry Pi required 44 hours.  
