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

I have see $n=3,k=1$,it is Classical inequalities see this :Prove inequality $\sqrt{\frac{2a}{b+a}} + \sqrt{\frac{2b}{c+b}} + \sqrt{\frac{2c}{a+c}} \leq 3$

when $n=4,k=2$ it is also Classical inequalities $$\sum_{cyc}\sqrt{\dfrac{a}{a+b+c}}\le\dfrac{4}{\sqrt{3}}$$

But general How to solve it?Thanks

math110
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  • The conjecture of $\lambda=\dfrac{n}{\sqrt{k+1}}$ may not be true since if you take $a_i=p^{-i}$ and let $p\rightarrow \infty$, the left tends to $n-k$. – Guangyi Zou Jul 18 '21 at 09:34

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