How to prove that $2\sqrt{3}$ is greater than $\pi$ Without calculator, how to prove that $2 \sqrt{3} > \pi$?
The level is baccalauréat grade.
I confirm it's not a school exercise at all, as I left school like 35 years ago.

By a geometric argument just consider a  circle inscribed in an hexagon and compare the two perimeters to obtain
$$3 \cdot \frac23 \sqrt 3 > \pi \cdot 1 \iff 2\sqrt 3>\pi$$

