<s>[INST] How prove $ \sqrt{2}+\sqrt{3}>\pi$? How prove that $ \sqrt{2}+\sqrt{3}>\pi$? Maybe some easy way?
 [/INST] This is just an improved version of The Great Seo's answer.
Since:
$$\sum_{n=1}^{+\infty}\frac{1}{n^2\binom{2n}{n}}=\frac{\pi^2}{18},\qquad\sum_{n=1}^{+\infty}\frac{1}{n^4\binom{2n}{n}}=\frac{17\,\pi^4}{3240}$$
you only need to check that:
$$\sum_{n=1}^{+\infty}\frac{\frac{3240}{17}-180\,n^2}{n^4\binom{2n}{n}}<-1$$
that is trivial since 
$$\sum_{n=1}^{3}\frac{\frac{3240}{17}-180\,n^2}{n^4\binom{2n}{n}}<-1$$
yet, and the extra terms are negative. As an alternative, since the archimedean approximation $\pi<\frac{22}{7}$ holds,
$$(\pi^2-5)^2 < \left(\left(\frac{22}{7}\right)^2-5\right)^2 = \frac{57121}{2401}<24.$$
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