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.


(Credit to David G. Stork for the image).
I am showing the area-based argument explicitly because it seems that the other answers rely on a perimeter-based argument (which I find unconvincing without a rigorous proof). In contrast, it is quite easy to conclude by simple inspection that the circumscribed hexagon has a larger area than the inscribed circle. 
The area of the circle is clearly simply $\pi$.
The hexagon can be decomposed into six congruent equilateral triangles. The height of each is $1$. The base can be computed with trigonometry as $(2)\tan\frac{\pi}{6} = \frac 2{\sqrt 3}$. Hence the area of a single triangle is $\frac 12 (1)(\frac 2{\sqrt 3})= \frac 1{\sqrt 3}$. The area of the hexagon is therefore $\frac 6{\sqrt 3} = 2\sqrt 3$.
This allows us to immediately conclude $2\sqrt 3> \pi$ as required.
