How to prove that $\sqrt{2}+\sqrt{3}>\pi$

Does someone know a other proofs (using properties of $\pi$) of following inequality:

$$\sqrt{2}+\sqrt{3}>\pi$$

First proof: the area of ​​regular 48-gon circumscribed to the unit circle is greater than $\pi$ and less than $\sqrt{2}+\sqrt{3}$.

Second proof: $\pi<\frac{355}{113}<\sqrt{2}+\sqrt{3}$

Any hints would be appreciated.

By evaluating the integral

$$\int_0^1 \frac{x^4 (1 - x)^4}{1 + x^2} = \frac {22} 7 - \pi$$

as is done here, we can show that $\frac{22} 7 > \pi$.

On the other hand, by squaring both sides and rearranging, it's fairly easy to show that $\sqrt{2} + \sqrt{3} > \frac{22} 7$.