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

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    $\begingroup$ Not sure that it helps, but it is equivalent to $(\pi^{2}-5)^{2} < 24.$ $\endgroup$ – Geoff Robinson Aug 1 '14 at 6:50
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    $\begingroup$ Approximating $\sqrt{2}$ and $\sqrt{3}$ up to three decimal places is very easy. Approximation of $\pi$ can be done using upper and lower Riemann sums of $2\sqrt{1 - x^2}$ function. $\endgroup$ – William Aug 1 '14 at 6:53
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    $\begingroup$ Some solutions here. $\endgroup$ – Ian Mateus Aug 1 '14 at 7:10
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Assuming known inequality $\pi<\frac{22}{7}$, it's easy to proceed by proving that $\frac{22}{7}<\sqrt{2}+\sqrt{3}$: $$\frac{484}{49}<5+2\sqrt6,$$ $$\frac{239}{49}<2\sqrt6,$$ $$\left(\frac{239}{98}\right)^2<\left(\frac{120}{49}\right)^2=\frac{14400}{2401}<6,$$ $$14400<14406.$$

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  • $\begingroup$ you are supposed to provide a succint proof that uses integral or some other estimates and not using 22/7. $\endgroup$ – DeepSea Aug 1 '14 at 7:00
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    $\begingroup$ Then it would have been helpful if you said so at the outset. $\endgroup$ – Harald Hanche-Olsen Aug 1 '14 at 7:01
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    $\begingroup$ @8pir Who said so? $\endgroup$ – martini Aug 1 '14 at 7:02
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    $\begingroup$ @8pir The OP never specified a specific way to prove. $\endgroup$ – Cookie Aug 1 '14 at 7:02
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    $\begingroup$ I think the main point of this question is how to approximate $\pi$. The result that $\pi < \frac{22}{7}$ require a bit of work. There is even a wikipedia article on it: en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80 $\endgroup$ – William Aug 1 '14 at 7:03
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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.

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    $\begingroup$ Isn't the last infinite sum $\frac17$ instead of $\frac18$? $\endgroup$ – User Aug 1 '14 at 7:24
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    $\begingroup$ @Matteo Oh, sorry. I edited. $\endgroup$ – The Great Seo Aug 1 '14 at 7:30
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Use your favorite method to show

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

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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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