This was asked but never answered at MSE, where it has a second open bounty.

Given fixed positive integers $n,k$, determine the minimal constant $\lambda = \lambda(n,k)$ for which the following inequality holds for any $a_1,a_2,...,a_n>0$ (taking indices mod $n$ if required): $$\sum_{i=1}^n\frac{a_i}{\sqrt{a_i^2+a_{i+1}^2+...+a_{i+k}^2}}\le \lambda$$

It seems that $\lambda=\dfrac{n}{\sqrt{k+1}}??$

I have seen for $n=3, k=1$, it is a classical inequality; see https://math.stackexchange.com/questions/1481348/prove-inequality-sqrt-frac2aba-sqrt-frac2bcb-sqrt-frac2c.

**Now I have solve when $n=3,4$ case:**

When $n=4, k=2$ it is also a classical inequality $$\sum_{cyc}\sqrt{\dfrac{a}{a+b+c}}\le\dfrac{4}{\sqrt{3}}$$

When $n=4,k=1$ it is also a classical inequality

For $n=4,k=3$,it is clear $$\sum_{cyc}\sqrt{\dfrac{a}{a+b+c+d}}\le 2$$ Because WLOG $a+b+c+d=1$,then use Cauchy-Schwarz inequality $$\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le \sqrt{4(a+b+c+d)}=2$$ But general How to solve it? Thanks