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

How prove that $ \sqrt{2}+\sqrt{3}>\pi$? Maybe some easy way?

Assuming known inequality $\pi<\frac{22}{7}$, it's easy to proceed by proving that $\frac{22}{7}<\sqrt{2}+\sqrt{3}$: $$\frac{484}{49}<5+2\sqrt6,$$ $$\frac{239}{49}<2\sqrt6,$$ $$\left(\frac{239}{98}\right)^2<\left(\frac{120}{49}\right)^2=\frac{14400}{2401}<6,$$ $$14400<14406.$$

Use this: $$\begin{align}{\pi^2\over6} &=\sum_{n=1}^\infty {1\over n^2} \\&=\sum_{n=1}^{10}\frac1{n^2}+\sum_{n=11}^\infty {1\over n^2} \\&\le\sum_{n=1}^{10}\frac1{n^2}+\sum_{n=11}^\infty {1\over(n-1)n} \\&=\sum_{n=1}^{10}\frac1{n^2}+\sum_{n=11}^\infty \left({1\over n-1}-\frac1n\right) \\&=\sum_{n=1}^{10}\frac1{n^2}+\frac1{10}. \end{align}$$

Thus we have $$(\pi^2-5)^2\le\left(\left(\sum_{n=1}^{10}\frac1{n^2}+\frac1{10}\right)\times6-5\right)^2=23.996\ldots<24,$$ and this completes the proof, as Geoff Robinson pointed out.

Use your favorite method to show

$$ \sqrt{2} > 1.414$$ $$ \sqrt{3} > 1.73$$ $$ \pi < 3.144$$

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, $$(\pi^2-5)^2 < \left(\left(\frac{22}{7}\right)^2-5\right)^2 = \frac{57121}{2401}<24.$$