A square inscribed inside of a circle with radius $1$ must have a perimeter $4\sqrt2$. A regular hexagon circumscribed about that same circle has a perimeter $4\sqrt3$. Since we know the circle’s circumference is $2\pi$, we can prove the inequality $2\sqrt2 < \pi<2\sqrt3$.
By averaging the the lower and upper bounds of that inequality, we can find an approximation for $\pi$: $$\pi \approx \sqrt2 + \sqrt3$$
Since $\sqrt2 + \sqrt3 \approx 3.14626$, and $\pi \approx 3.14159$, it is clear that $\pi$ is less than $\sqrt2 + \sqrt3$. But how can we prove that without knowing the value of $\pi$?
