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

Question

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

Not sure that it helps, but it is equivalent to $(\pi^{2}-5)^{2} < 24.$ — Geoff Robinson, Aug 01 '14 at 06:50
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. — William, Aug 01 '14 at 06:53

score 10 · Accepted Answer · answered Aug 01 '14 at 06:59

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

answered Aug 01 '14 at 06:59

CuriousGuest

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you are supposed to provide a succint proof that uses integral or some other estimates and not using 22/7. – DeepSea Aug 01 '14 at 07:00
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Then it would have been helpful if you said so at the outset. – Harald Hanche-Olsen Aug 01 '14 at 07:01
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@8pir Who said so? – martini Aug 01 '14 at 07:02
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@8pir The OP never specified a specific way to prove. – Cookie Aug 01 '14 at 07:02
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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: http://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80 – William Aug 01 '14 at 07:03
See also http://math.stackexchange.com/questions/701822/geometric-explanation-of-sqrt-2-sqrt-3-approx-pi. – lhf Aug 01 '14 at 11:51
I'm not sure about the inequality proof since you use the assumption that it is true throughout the proof to get your result. I'd say use a proof by contradiction and then prove it false. – Ryan Dougherty Aug 01 '14 at 15:21

Jaehyeon Seo · Answer 2 · 2014-08-01T07:39:40.257

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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.

edited Aug 01 '14 at 07:39

answered Aug 01 '14 at 07:12

Jaehyeon Seo

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Isn't the last infinite sum $\frac17$ instead of $\frac18$? – User Aug 01 '14 at 07:24
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@Matteo Oh, sorry. I edited. – Jaehyeon Seo Aug 01 '14 at 07:30
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The first equality is nontrivial. Proving it would make this answer much longer. – John Bentin May 21 '20 at 19:06

score 1 · Answer 3 · answered Aug 01 '14 at 08:51

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Use your favorite method to show

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

answered Aug 01 '14 at 08:51

score 1 · Answer 4 · answered Aug 01 '14 at 11:27

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

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

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