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

[anemone]

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.

 

[Euge]

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.

Spoiler

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}} $.

 

[anemone]

Spoiler

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!