<s>[INST] 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 $(1)$.
 [/INST] If we consider the Beuker-like integral
$$ 0<\int_{0}^{1}\frac{x^8(1-x)^8}{1+x^2}\,dx = 4\pi-\frac{188684}{15015} $$
we get, through partial fraction decomposition and few operations in $\mathbb{Q}$,
$$ \pi > \frac{47171}{15015} > 3.14159.$$

Inspired by Professor Vector's brilliant approach, I am adding a further approach.
By the Shafer-Fink inequality we have $\arctan(x)>\frac{3x}{1+2\sqrt{1+x^2}}$ for any $x>0$, hence by evaluating both sides at $x=\frac{1}{\sqrt{3}}$ we get $\pi>\frac{18}{13}(4-\sqrt{3})=3.140237\ldots$ A refinement of the previous inequality is
$$\forall x>0,\qquad \arctan(x)>\frac{6x}{1+\sqrt{1+x^2}+2\sqrt{2}\sqrt{1+x^2+\sqrt{1+x^2}}} $$
and the evaluation at $x=\frac{1}{\sqrt{3}}$ produces the sharper bound
$$ \pi > \frac{36}{2+\sqrt{3}+4 \sqrt{2+\sqrt{3}}} > 3.1415.$$

Yet another approach. The inequality $\arctan(x)>\frac{5x(21+11x^2)}{105+90x^2+9x^4}$ for any $x\in(0,1)$ comes from the Gauss continued fraction / the Padé approximants for the arctangent function. By replacing $x$ with $\frac{x}{1+\sqrt{1+x^2}}$, then evaluating at $x=\frac{1}{\sqrt{3}}$, we get the nice and tight approximation:
$$ \pi > \color{blue}{\frac{5}{601}\left(944-327\sqrt{3}\right)}>3.141592.$$

There is also a nice geometric argument leading to an acceleration of Vieta's formula. Let $PQ$ be a side of a regular $n$-agon inscribed in a unit circle centered at $O$. Let $M$ be the midpoint of the minor arc $PQ$. We may consider the unique parabola through $P,M,Q$ and approximate the area of the circle sector delimited by $P,O,Q$ through $[POQ]$ plus the area of a parabolic segment, $\frac{4}{3}[PMQ]:$

It follows that if $A_n$ is the area of the inscribed $2^n$-agon, we have
$$ \pi \geq A_n+\frac{4}{3}\left(A_{n+1}-A_n\right) = \frac{4}{3}A_{n+1}-\frac{1}{3}A_n $$
with
$$ A_n = 2^{n-1} \sin\frac{\pi}{2^{n-1}} $$
and $\{A_n\}_{n\geq 2}$ being computable through a simple recursion,
$$ A_{n+1} = 2^n \sqrt{\frac{1-\sqrt{1-\frac{A_n^2}{4^{2n-2}}}}{2}}$$
where
$$ \frac{4}{3}A_5-\frac{1}{3}A_4 = \frac{4}{3}\left[8\sqrt{2-\sqrt{2+\sqrt{2}}}-\sqrt{2-\sqrt{2}}\right]=\color{green}{3.141}44\ldots $$
By considering a $12$-agon and a $24$-agon we get the simpler
$$ \pi > 4\sqrt{2}(\sqrt{3}-1)-1 = \color{green}{3.141}10\ldots $$
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