<s>[INST] Is there an integral that proves $\pi > 333/106$? The following integral,
$$ \int_0^1 \frac{x^4(1-x)^4}{x^2 + 1} \mathrm{d}x = \frac{22}{7} - \pi $$
is clearly positive, which proves that $\pi < 22/7$.
Is there a similar integral which proves $\pi > 333/106$?
 [/INST] From the relationship 
$$\frac{333}{106}=\frac{377-2·22}{120-2·7}$$
and integrals
$$\pi = \frac{22}{7} - \int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx$$
and
$$\pi = \frac{377}{120} - \frac{1}{2}\int_0^1 \frac{x^5(1-x)^6}{1+x^2}dx$$
we obtain
$$\int_0^1 \frac{14x^4(1-x)^4-60x^5(1-x)^6}{1+x^2}dx = \int_0^1 \frac{2x^4(1-x)^4(7-30x(1-x)^2)}{1+x^2}dx = \int_0^1 \frac{2x^4(1-x)^4(7-30x+60x^2-30x^3)}{1+x^2}dx = 106\pi-333$$
Therefore,

$$\frac{1}{53} \int_0^1 \frac{x^4(1-x)^4(7-30x+60x^2-30x^3)}{1+x^2}dx = \pi-\frac{333}{106}$$

The numerator can be shown to be nonnegative for $0\leq x\leq 1$.
In fact, we can get a smaller numerator that is still nonnegative, which leads to a closer approximation to $\pi$

$$\pi=\frac{21991}{7000} +\frac{1}{50} \int_0^1 \frac{x^4 (1 - x)^4 \left(4 - 27 x (1 - x)^2\right)}{\left(1 + x^2\right)} dx$$

Here the numerator has been adjusted to have a double zero in $(0,1)$ while not crossing the axis WA link
The resulting fraction is
$$\frac{21991}{7000} = \frac{22}{7}-\frac{9}{7000}$$
which allows for writing the double inequality
$$\frac{22}{7}-\frac{9}{7000}<\pi<\frac{22}{7}$$
or, equivalently, an upper bound for the error in Archimedes' approximation
$$\frac{22}{7}-\pi<\frac{9}{7000}$$
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