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 $(1)$.

Equation (2) in the question is the result of plugging $x=1$ in the expansion for the arctangent, using
$$\frac{\pi}{4}=\tan^{-1}(1)$$
Instead, from
$$\frac{\pi}{6}=\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)$$
a series with faster convergence is obtained. Taking six terms,
$$\pi>2\sqrt{3}\left(1-\frac{1}{3·3}+\frac{1}{5·3^2}-\frac{1}{7·3^3}+\frac{1}{9·3^4}-\frac{1}{11·3^5}\right)=\frac{509024\sqrt{3}}{280665}>3.141$$
Four terms from
$$\frac{\pi}{8}=\tan^{-1}\left(\sqrt{2}-1\right),$$
give
$$\pi> 8\left(\sqrt{2}-1-\frac{(\sqrt{2}-1)^3}{3}+\frac{(\sqrt{2}-1)^5}{5}-\frac{(\sqrt{2}-1)^7}{7}\right)=\frac{32}{105}\left(716-499\sqrt{2}\right)>3.141,$$
while two terms from
$$ \frac{\pi}{12}=\sin^{-1}\left( \frac{\sqrt{3}-1}{2\sqrt{2}} \right)$$
lead to
$$\pi > 12\left( \frac{\sqrt{3}-1}{2\sqrt{2}} + \frac{1}{6} \left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)^3 \right) = \frac{27\sqrt{3}-29}{4\sqrt{2}}>3.14$$
