Prove that for every natural \(n\), it is inequality:

\(\frac { 1 }{ 2 } \times \frac { 3 }{ 4 } \times \frac { 5 }{ 6 } ...\frac { 2n-1 }{ 2n } \le \frac { 1 }{ \sqrt { 3n+1 } } \)

Prove that for every natural \(n\), it is inequality:

\(\frac { 1 }{ 2 } \times \frac { 3 }{ 4 } \times \frac { 5 }{ 6 } ...\frac { 2n-1 }{ 2n } \le \frac { 1 }{ \sqrt { 3n+1 } } \)

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TopNewestI am not sure what exactly I am supposed to do here, even though I have been on Brilliant for a year I haven't been regular. So, I hope I am doing the right thing.

\(P(1)\) is true. Assume \(P(n)\) to be true.

\[P\left(n + 1\right) \implies \frac 1 2 \dots \frac {2n-1}{2n} \cdot \frac{2n + 1}{2n + 2} \leq \frac{1}{\sqrt{3n + 4}}\] From \(P(n)\), \[\frac{1}{\sqrt{3n + 1}} \cdot \frac{2n + 1}{2n + 2} \leq \frac{1}{\sqrt{3n + 4}}\] After squaring and cross-multiplying, \[\implies \left( 3n + 4 \right)\left(4n^2 + 4n + 1\right) \leq 4\left(n^2 + 2n +1\right)\left(3n + 1\right)\]

I hope you can draw the conclusion from here.

Note: This is actually a very standard problem and can be found in almost every math contest book (In case, anyone is interested.) – Ishaan Singh · 3 years, 2 months ago

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– Jon Haussmann · 3 years, 2 months ago

What is \(4(n^2 + 2n + 1)(3n + 1) - (4n^2 + 4n + 1)(3n + 4)\)? You should get something very simple.Log in to reply

First of all, we can prove that "\(\frac{2n+1}{2n+2} \leq \frac{\sqrt{3n+1}}{\sqrt{3n+4}}\)" is true.

Prove) Expansion the above inequality, we get \( (4n^2+4n+1)(3n+4) \leq (4n^2+8n+4)(3n+1) \). clean up this inequality, it become \( 0 \leq n \). this is alway true when n is natural number.

if P(n) to be true and "\(\frac{2n+1}{2n+2} \leq \frac{\sqrt{3n+1}}{\sqrt{3n+4}}\)" is also true, then \(\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \times \dots \times \frac{2n-1}{2n} \times \frac{2n+1}{2n+2} \leq \frac{1}{\sqrt{3n+1}} \times \frac{\sqrt{3n+1}}{\sqrt{3n+4}} \)". so P(n+1) is also true.

P(1) is true, If P(n) is true, P(n+1) also true. So for every n, this inequality is true. – Sung Moo Hong · 3 years, 2 months ago

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Nice question.

Edit: There is a easy proof for a looser bound go \( \frac{1} { \sqrt{2n+1} } \) that doesn't use induction.

Edit: The RHS can be tightened to \( \frac{ 1 } { \sqrt { \alpha n+ \beta} } \) for some constant \( \alpha < \pi \) and \( \beta \). However, this requires more work. – Calvin Lin Staff · 3 years, 2 months ago

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Incidentally, \[\frac{1}{2} \cdot \frac{3}{4} \dotsm \frac{2n - 1}{2n} = \frac{(2n)!}{4^n (n!)^2}.\] By Stirling's approximation, \[\frac{(2n)!}{4^n (n!)^2} \approx \frac{1}{\sqrt{\pi n}}.\] – Jon Haussmann · 3 years, 2 months ago

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Indeed, I wanted a coefficient of \( \pi \) instead. – Calvin Lin Staff · 3 years, 2 months ago

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