How to prove that $ \Bigg(\sum^{n}\limits_{k=1}\sqrt{\frac{k-\sqrt{k^{2}-1}}{\sqrt{k(k+1)}}}\Bigg)^{2} \le n\sqrt{\frac{n}{n+1}}$ for $n\ge1$

Question

I need to prove Prove the inequality $$ \Bigg(\sum^{n}_{k=1}\sqrt{\frac{k-\sqrt{k^{2}-1}}{\sqrt{k(k+1)}}}\Bigg)^{2} \le n\sqrt{\frac{n}{n+1}}, $$ where $n$ is a positive integer.

Equivalently $$\sum^{n}_{k=1}\sqrt{\frac{k-\sqrt{k^{2}-1}}{\sqrt{k(k+1)}}} \le \sqrt[4]{\frac{n^3}{n+1}}, $$

Then proceed by induction assuming it is true for $n-1$ we have

$$\sum^{n}_{k=1}\sqrt{\frac{k-\sqrt{k^{2}-1}}{\sqrt{k(k+1)}}} \le \sqrt[4]{\frac{(n-1)^3}{n}} + \sqrt{\frac{n-\sqrt{n^{2}-1}}{\sqrt{n(n+1)}}} $$

Then one need to show that the last term is not greater than $$\sqrt[4]{\frac{n^3}{n+1}}$$ which is not an obvious task.

Question; Is there an alternative way of proving this without appealing induction or can anyone help to prove this last step?

Answer 1

It is a combination of Cauchy-Schwarz inequality and a telescope sum.

According to Cauchy-Schwarz inequality:

$$\Bigg(\sum^{n}_{k=1}\sqrt{\frac{k-\sqrt{k^{2}-1}}{\sqrt{k(k+1)}}}\Bigg)^{2} \le \left( \sum_{k=1}^n 1^2 \right)\left( \sum_{k=1}^n \frac{k-\sqrt{k^{2}-1}}{\sqrt{k(k+1)}} \right) = n \cdot \left( \sum_{k=1}^n \frac{k-\sqrt{k^{2}-1}}{\sqrt{k(k+1)}} \right)$$

The sum on the right-hand side can now be transformed into a telescoping sum:

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

Done.