A simple proof related to pi
The first published statement of this result was in 1971 by Dalzell. It was also presented as a problem in the 1968 Putnam Competition, and later it also came in the JEE-Adv. The problem was to compute the integral,
\[\int_0^1 \frac{x^4(1 - x)^4}{1 + x^2}\:dx.\]I encourage readers to try it at first.
Solution:
\[\begin{aligned} \int_0^1 \frac{x^4(1 - x)^4}{1 + x^2}\:dx &= \int_0^1 x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac{4}{1 + x^2} \:dx \\ &= \left[\frac{x^7}{7} - \frac{2x^6}{3} + x^5 - \frac{4x^3}{3} + 4x - 4\arctan{x}\right]_0^1 \\ &= \frac{1}{7} - \frac{2}{3} + 1 - \frac{4}{3} + 4 - \pi \\ &= \frac{22}{7} - \pi. \end{aligned}\]Now observe one thing:
\[\frac{x^4(1 - x)^4}{1 + x^2} > 0 \implies \int_0^1 \frac{x^4(1 - x)^4}{1 + x^2}\:dx > 0.\]Hence,
\[\frac{22}{7} > \pi.\]About the author
Sougata Panda is a student of the Chennai Mathematical Institute. This particular article placed third in our Article Writing Contest, 2022.