P103.4.2.2018.2.2 Given positive integers $n, k$ such that $n\ge 4k$, find the minimal value $\lambda=\lambda(n,k)$ such that for any positive reals $a_1,a_2,\ldots,a_n$, we have
\[ \sum\limits_{i=1}^{n} {\frac{{a}_{i}}{\sqrt{{a}_{i}^{2}+{a}_{{i}+{1}}^{2}+{\cdots}{{+}}{a}_{{i}{+}{k}}^{2}}}} \le \lambda\] 
Where  $a_{n+i}=a_i,i=1,2,\ldots,k$

Solution

With the condition $n \geq 4k$, we say that the answer $\lambda=\lambda(n,k)=n-k$. The following is the clue.
We need a lemma:
For positive reals $a,b,c,d$
, $$\sum_{cyc} \frac{a}{\sqrt{a^2+b^2}} \leq 3$$
the proof for the lemma is not hard, one can have both sides squared and apply Cauchy-Schwarz inequality. $\blacksquare$
Now back to the main problem, for $n=4k$ we can dissect $4k$ variables to $k$ groups: $a_i,a_{i+k},a_{i+2k},a_{i+3k}, 1 \leq i \leq k$. Ignore $a_{i+1}^2+...+a_{i+k-1}^2$ and apply the lemma, we are done.
To end the proof, we use induction on $n$. Consider the maximal $a_i$ and delete it, the rest $a_is$ still form a cycle and by the maximality of $a_i$ the value doesn't increase. What we deleted is at most 1, by induction hypothesis, we are done. $\blacksquare$
To see the $\lambda$ is the best constant, consider $a_is$ in isometric series, and the ratio is sufficiently large.

Let $(n,k)=(3,1),$
Suppose that $ a, b, c$ are positive real numbers, prove that
$$\sqrt{\frac {a}{a+ b}}+\sqrt{\frac {b}{b+ c}}+\sqrt{\frac {c}{c+ a}}\leq\frac {3\sqrt {2}}{2}$$
When$(n,k)=(4,2),$ Hanjingjun have:
Suppose that $ a, b, c,d$ are positive real numbers, prove that
$$\sqrt{\frac {a}{a+ b+c}}+\sqrt{\frac {b}{b+ c+d}}+\sqrt{\frac {c}{c+d+ a}}+\sqrt{\frac {d}{d+a+b}}\leq\frac {4\sqrt {3}}{3}$$

Guess of Hanjingjun:
Suppose that $ a, b,c,d,e$ are positive real numbers, prove that

$$\sum_{cyc}\sqrt {\frac {a}{a+ b+c}} \leq 3$$
$$\sum_{cyc}\sqrt {\frac {a}{a+ b+c+d}} \leq \frac {5}{2}$$