Simplest way to get the lower bound $\pi > 3.14$
From the elementary inequality $$\frac{\sin x}x\le\frac{2+\cos x}3,$$ we get with $x=\pi/6$ easily $\pi\ge\frac{18}{4+\sqrt{3}}=3.1402\ldots$
Proof of the inequality (elementary, though not obvious): let $$f(x)=\frac{\sin x}{x(2+\cos x)}.$$ In order to prove $f(x)\le\lim_{x\to+0}f(x)$, we prove $f(x)\le f(x/2)$. That follows from $$f(x)=f(x/2)\,\frac{(2+\cos x/2)\cos x/2}{1+2\cos^2 x/2},$$ since with $c=\cos x/2$, we have $$\frac{(2+c)c}{1+2c^2}=1-\frac{(1-c)^2}{1+2c^2}\le1.$$
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 $$ \pi > 4\sqrt{2}(\sqrt{3}-1)-1 = \color{green}{3.141}10\ldots $$
Take Machin's formula: $$\pi=16\tan^{-1}\frac15-4\tan^{-1}\frac1{239}$$ Expand the arctangents into their Taylor series: $$\pi=16\left(\color{blue}{\frac15-\frac1{5^3×3}}+\frac1{5^5×5}-\dots\right)-4\left(\color{blue}{\frac1{239}}-\frac1{239^3×3}+\dots\right)$$ $$=\color{blue}{\frac{16}5-\frac{16}{375}-\frac4{239}}+\delta$$ $$=\color{blue}{3.140596\dots}+\delta$$ with $0<\delta<\frac{16}{5^5×5}+\frac4{239^3×3}$ since the two series are alternating with term magnitudes strictly decreasing. This proves $\pi>3.140596\dots>3.14$.