How prove $ \sqrt{2}+\sqrt{3}>\pi$? How prove that $ \sqrt{2}+\sqrt{3}>\pi$? Maybe some easy way?
 A: 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.
A: Assuming known inequality $\pi<\frac{22}{7}$, it's easy to proceed by proving that $\frac{22}{7}<\sqrt{2}+\sqrt{3}$:
$$\frac{484}{49}<5+2\sqrt6,$$
$$\frac{239}{49}<2\sqrt6,$$
$$\left(\frac{239}{98}\right)^2<\left(\frac{120}{49}\right)^2=\frac{14400}{2401}<6,$$
$$14400<14406.$$
A: Use your favorite method to show
$$ \sqrt{2} > 1.414$$
$$ \sqrt{3} > 1.73$$
$$ \pi < 3.144$$
A: 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,
$$(\pi^2-5)^2 < \left(\left(\frac{22}{7}\right)^2-5\right)^2 = \frac{57121}{2401}<24.$$
