MHB Can You Solve the Summation of Series Challenge Using Cauchy-Schwarz Inequality?

AI Thread Summary
The challenge involves proving the inequality $\left(\sum_{k=1}^{n} \sqrt{\dfrac{k-\sqrt{k^2-1}}{\sqrt{k(k+1)}}}\right)^2\le n\sqrt{\dfrac{n}{n+1}}$ for positive integers n. The Cauchy-Schwarz inequality is identified as the essential tool for solving this problem. Participants discuss the application of this inequality to derive the necessary proof. The conversation emphasizes the importance of understanding the underlying mathematical principles. Engaging with this challenge enhances problem-solving skills in mathematical inequalities.
anemone
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Prove that $\displaystyle\left(\sum_{k=1}^{n} \sqrt{\dfrac{k-\sqrt{k^2-1}}{\sqrt{k(k+1)}}}\right)^2\le n\sqrt{\dfrac{n}{n+1}}$, where $n$ is a positive integer.
 
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anemone said:
Prove that $\displaystyle\left(\sum_{k=1}^{n} \sqrt{\dfrac{k-\sqrt{k^2-1}}{\sqrt{k(k+1)}}}\right)^2\le n\sqrt{\dfrac{n}{n+1}}$, where $n$ is a positive integer.

By the Cauchy-Schwarz inequality,

$\displaystyle\left(\sum_{k=1}^{n} \sqrt{\dfrac{k-\sqrt{k^2-1}}{\sqrt{k(k+1)}}}\right)^2 \le n \sum_{k=1}^{n} \dfrac{k-\sqrt{k^2-1}}{\sqrt{k(k+1)}} = n \sum_{k = 1}^n \left(\sqrt{\frac{k}{k+1}} - \sqrt{\frac{k-1}{k}}\right) = n\sqrt{\frac{n}{n+1}} $.
 
Euge said:
By the Cauchy-Schwarz inequality,

$\displaystyle\left(\sum_{k=1}^{n} \sqrt{\dfrac{k-\sqrt{k^2-1}}{\sqrt{k(k+1)}}}\right)^2 \le n \sum_{k=1}^{n} \dfrac{k-\sqrt{k^2-1}}{\sqrt{k(k+1)}} = n \sum_{k = 1}^n \left(\sqrt{\frac{k}{k+1}} - \sqrt{\frac{k-1}{k}}\right) = n\sqrt{\frac{n}{n+1}} $.

Thanks for participating, Euge!

Yes, the key to unlock this problem is the Cauchy-Schwarz inequality. Good job, Euge!
 
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