Geometric proof for inequality While on AOPS, I saw this interesting problem. I was wondering how many different approaches could be used to tackle the problem. 
In other words I am looking for interesting and unique ways to solve the question: 

Show geometrically that $\sqrt{2}+\sqrt{3}>\pi$


The tools we need to do this were provided to us in 1654 by Christiaan Huygens in his De Circuli Magnitudine Inventa. I'll rely on his Proposition IX, as well as this derivation at Wikipedia, both of which are entirely geometrical in nature.
The former states that the circumference of a circle is less than two thirds the perimeter of an inscribed regular polygon plus one third the perimeter of a circumscribed regular polygon with the same number of side, while the latter gives the perimeter of an inscribed regular polygon as the geometric mean of the perimeters of a circumscribed regular polygon with the same number of sides and an inscribed regular polygon with half the number of sides, and the perimeter of a circumscribed regular polygon as the harmonic mean of inscribed and circumscribed regular polygons with half the number of sides.
Thus, denoting the perimeter of a regular $n$-gon inscribed into the unit circle by $u_n$ and the perimeter of a regular $n$-gon circumscribed around the unit circle by $U_n$, we have:
$$u^2_{2n}=u_nU_{2n}\;,$$
$$\frac2{U_{2n}}=\frac1{u_n}+\frac1{U_n}\;,$$
$$2u_n+U_n\gt6\pi\;.$$
If we seed the recurrence with regular hexagons and write $\sigma=2-\sqrt3$, we obtain
$$
\begin{array}{c|c|l}
n&u_n&U_n&2u_n+U_n\\
\hline\\
6&6&4\sqrt3&12+4\sqrt3\\
12&12\sqrt{\sigma}&24\sigma&24(\sqrt\sigma+\sigma)
\end{array}
$$
Thus $4(\sqrt\sigma+\sigma)\gt\pi$. A couple of elementary squaring steps yield $\sqrt2+\sqrt3\gt4(\sqrt\sigma+\sigma)$, and thus $\sqrt2+\sqrt3\gt\pi$.
