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 $(1)$.
 A: Another simple method is to use the Newton-Raphson method to find the zero of the function $\sin(x)$ at $x = \pi$ using the initial guess of $x = 3$. The fact that at $x = \pi$ the function $\sin(x)$ has a point of inflexion, means that convergence to $x = \pi$ will happen in an alternating way, therefore you obtain rigorous upper and lower bounds. So, the sequence defined by the recurrence:
$$x_{n+1} = x_n - \tan(x_n)$$ 
and $x_0 = 3$ converges to $\pi$. We have $x_1 = 3.1425\cdots$, $x_2 = 3.1415926533\cdots$, and therefore $\pi > 3.1415926533$. Now, we need to compute $x_2$ here without using $\pi$ itself, we can use the series expansions of $\sin(x)$ and $\cos(x)$ around $x = 0$. These are both alternating series, so we can easily get to an accurate bound on $x_1$ and from that we can get to an accurate bound on $x_2$, the lower bound of which then yields an accurate lower bound for $\pi$.
A: 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 $$\frac{(2+c)c}{1+2c^2}=1-\frac{(1-c)^2}{1+2c^2}\le1.$$
A: A very easily understandable way is to use regular polygons inscribed into a circle of radius $1$. If $a_n$ is the length of the side of the $n$-gon, then $$a^2_{2n} = 2-\sqrt{4-a^2_n}\text{.}$$ 
A lower bound on $a_n$ implies a lower bound on $a_{2n}$, we can thus round down intermediary results.
Four applications of the formula with 4 to 5 significant digits are enought to establish the bound of $3.14$.
$$\begin{array}{rl}
a^2_6 =& 1\\
a^2_{12} >& 0.2679 \\
a^2_{24} >& 0.06813 \\
a^2_{48} >& 0.017105 \\
a^2_{96} >& 0.004280 \\
\end{array}$$
Leading to $a_{96} > 0.06542$ and $\pi > 48a_{96} > 3.14$.
A: From $$\pi=\sum_{k=0}^{+ \infty}{\dfrac {1}{16^k}} \cdot \dfrac {120k^2+151k+47}{512k^4+1024k^3+712k^2+194k+15}$$ you obtain your inequality by summing just the first two terms.
A: 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
$$ \pi > 4\sqrt{2}(\sqrt{3}-1)-1 = \color{green}{3.141}10\ldots $$
A: 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 $\pi>3.140596\dots>3.14$.
A: 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
$$\pi > 12\left( \frac{\sqrt{3}-1}{2\sqrt{2}} + \frac{1}{6} \left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)^3 \right) = \frac{27\sqrt{3}-29}{4\sqrt{2}}>3.14$$
A: From series
$$\pi^6=960\sum_{k=0}^\infty \frac{1}{(2k+1)^6},$$
$$\pi^6>960$$
But $3.14^6<960$, so $3.14^6<\pi^6$ and therefore $3.14<\pi$.
A: For modest requirements of accuracy, we can use a quite basic and easy to understand convergence acceleration technique to transform the Gregory/Leibniz series into something that yields the result with little computation.
If we have an alternating series
$$\sum_{n = 0}^{\infty} (-1)^n a_n$$
where the $a_n$ are slowly converging to $0$, then it takes little guessing to believe that the arithmetic mean of two successive partial sums of the series is a much better approximation to the value of the series than either or the two partial sums. We can write the mean as
$$\sum_{n = 0}^{m-1} (-1)^n a_n + \frac{(-1)^m a_m}{2} = \frac{a_0}{2} + \frac{1}{2} \sum_{n = 0}^{m-1} (-1)^n(a_n - a_{n+1})$$
and see that if $\bigl(a_n - a_{n+1}\bigr)_{n\in \mathbb{N}}$ is a decreasing sequence we immediately have an error bound of $\frac{1}{2}(a_{m} - a_{m+1})$ for this arithmetic mean. If $(a_n)$ is slowly converging, then this is much smaller than either of $a_m$ and $a_{m+1}$. For nice $(a_n)$, this can be iterated.
If we apply that to $a_n = \frac{1}{2n+1}$, we find $a_n - a_{n+1} = \frac{2}{(2n+1)(2n+3)}$, which is decreasing - and still converges slowly to $0$, so we get a significant improvement from then applying the method again. We get
\begin{align}
\frac{\pi}{4} &= \sum_{n = 0}^{\infty} \frac{(-1)^n}{2n+1} \\
&= \frac{1}{2} + \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)(2n+3)} \\
&= \frac{1}{2} + \frac{1}{6} + 2\sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)(2n+3)(2n+5)} \\
&= \frac{1}{2} + \frac{1}{6} + \frac{1}{15} + 6\sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)(2n+3)(2n+5)(2n+7)}
\end{align}
and so on, each series converging faster than the previous. Since the terms get more complicated, it is computationally - at least when done by hand - better to start accelerating not right at the beginning of the series, but rather compute a partial sum of the original series, and then use acceleration only on the remainder. Let's say that starting the remainder at $n = 4$ wouldn't need an unreasonable amount of computation, then we get
\begin{align}
\frac{\pi}{4} &= 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \sum_{n = 4}^{\infty} \frac{(-1)^n}{2n+1} \\
&= 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{18} + \sum_{n = 4}^{\infty} \frac{(-1)^n}{(2n+1)(2n+3)} \\
&= 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{18} + \frac{1}{198} + 2\sum_{n = 4}^{\infty} \frac{(-1)^n}{(2n+1)(2n+3)(2n+5)} \\
&= 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{18} + \frac{1}{198} + \frac{1}{1287} + 6\sum_{n = 4}^{\infty} \frac{(-1)^n}{(2n+1)(2n+3)(2n+5)(2n+7)} \\
&> 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{18} + \frac{1}{198} + \frac{1}{1287} \\
&> 0.78519
\end{align}
and thus $\pi > 3.14076$.
Of course Machin-like formulæ are far more efficient. But it's not too terrible.
A: 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.  
A: 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 $\pi$) 
