<s>[INST] 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 $n\ge1$ 
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?


 [/INST] 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.
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