Note that

\[{ \int _0 ^{1} \frac{1}{1+x ^2} \, dx = \frac{\pi}{4} . }\]

This suggests studying inequalities of form

\[{ \boxed{ \int _0 ^{1} \frac{P(x) ^2}{1 + x ^2} \, dx > 0 } }\]

where ${ P(x) }$ is a polynomial, to get bounds on ${ \pi . }$

Ideally, we want ${ P(x) ^2 }$ to leave a constant remainder on dividing by ${ (1 + x ^2) . }$

Let’s write

\[{ P(x) ^2 = (1 + x ^2) Q(x) + R(x) . }\]

Putting ${ x = i }$ we have

\[{ P(x) ^2 = (1 + x ^2) Q(x) + P(i) ^2 . }\]

We want ${ P(i) ^2 }$ to be real.

Suppose for now we want an upper bound on ${ \pi . }$ Hence we can set ${ P(i) }$ to be purely imaginary, to get an upper bound on ${ \pi . }$ (Note that if ${ P(i) }$ is purely imaginary, we have the remainder ${ P(i) ^2 < 0 }$).

We can try

\[{ P(x) = a _4 x ^4 + \ldots + a _1 x + a _0 . }\]

By the ${ P(i) }$ purely imaginary condition we have

\[{ a _4 + a _0 = a _2 . }\]

For simplicity we can look at ${ a _0 = 0 . }$ Hence ${ a _4 = a _2 . }$

By scaling WLOG ${ a _4 = a _2 = 1 . }$

Hence consider

\[{ P(x) = x ^4 + a _3 x ^3 + x ^2 . }\]

Note that

\[{ {\begin{aligned} &\, ( x ^4 + a _3 x ^3 + x ^2 ) ^2 \\ = &\, (1 + x ^2) (\text{Quotient}) + P(i) ^2 \\ = &\, (1+x ^2) (\text{Quotient}) - a _3 ^2 . \end{aligned}} }\]

For elegance ${ a _3 = (-2) }$ ensures

\[{ \boxed{ P(x) = x ^2 (1-x) ^2 } }\]

and

\[{ P(x) ^2 = (1+x ^2) (\text{Quotient}) - 4 . }\]

It turns out by using this ${ P(x) }$ in

\[{ \int _0 ^{1} \frac{P(x) ^2}{1 + x ^2} \, dx > 0 }\]

we get

\[{ \frac{22}{7} - \pi > 0 . }\]

Plotting this ${ P(x) ^2 / (1 + x ^2) }$ on ${ [0,1] }$ shows how small the LHS practically is.

P.S. This argument also explains why using

\[{ P(x) = x ^{4k+2} (1-x) ^{4k+2} }\]

give better and better upper bounds for ${ \pi . }$ (Note that ${ P(i) }$ is purely imaginary, hence the remainder ${ P(i) ^2 < 0}$).

Link to MathOverflow answer by Geoff Robinson: Link.

The cases

\[{ P(x) = x ^{4k} (1 - x) ^{4k} }\]

give better and better lower bounds for ${ \pi . }$ (Note that ${ P(i) }$ is purely real, hence the remainder ${ P(i) ^2 \geq 0 }$).

For eg, the ${ k = 1 }$ case, as mentioned in Noam Elkies’ MathOverflow post here, gives a lower bound on ${ \pi }$:

\[{ \int_0^1 (x-x^2)^8 \frac{dx}{1+x^2} = 4\pi - \frac{188684}{15015} > 0 . }\]