smy2012 wrote: 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 $a_{n+i}=a_i,i=1,2,\ldots,k$ See also here https://artofproblemsolving.com/community/c6h1572307p9659757

smy2012 wrote: 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 $a_{n+i}=a_i,i=1,2,\ldots,k$ When$(n,k)=(3,1),$ it is CWMO 2004, problem 8: 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 $$\sum_{cyc}\sqrt {\frac {a}{a+ b+c}} \leq 3$$$$\sum_{cyc}\sqrt {\frac {a}{a+ b+c+d}} \leq \frac {5}{2}$$

Any sol??

答案 $n-k$. 构造 对 $1\leq i\leq n$, 取 $a_i=q^i$, 其中 $0<q<1$. 令 $q\to 0$. 则 当 $1\leq i\leq n-k$ 时, $\frac{{a}_{i}}{\sqrt{{a}_{i}^{2}+{a}_{{i}+{1}}^{2}+{\cdots}{{+}}{a}_{{i}{+}{k}}^{2}}}=\frac{1}{\sqrt{1+q+\cdots +q^k}}\to 1$. 当 $n-k<i\leq n$ 时, $\frac{{a}_{i}}{\sqrt{{a}_{i}^{2}+{a}_{{i}+{1}}^{2}+{\cdots}{{+}}{a}_{{i}{+}{k}}^{2}}}=\frac{q^{i-1}}{\sqrt{q^{2(i-1)}+\cdots +q^{2(n-1)}+1+q^2+\cdots +q^{2(i+k-n-1)}}}\to 0$. 此时 $\sum\limits_{i=1}^n\frac{{a}_{i}}{\sqrt{{a}_{i}^{2}+{a}_{{i}+{1}}^{2}+{\cdots}{{+}}{a}_{{i}{+}{k}}^{2}}}\to n-k$, 因此 $\lambda\geqslant n-k$. 证明 下面我们证明, 总有 $\sum\limits_{i=1}^n\frac{{a}_{i}}{\sqrt{{a}_{i}^{2}+{a}_{{i}+{1}}^{2}+{\cdots}{{+}}{a}_{{i}{+}{k}}^{2}}}\leqslant n-k$. 我们先证明 $n=4$, $k=1$ 时结论成立. 将两边分别平方, 则只需证明 $$\frac{a_1^2}{a_1^2+a_2^2}+\frac{a_2^2}{a_2^2+a_3^2}+\frac{a_3^2}{a_3^2+a_4^2}+\frac{a_4^2}{a_4^2+a_1^2}+\frac{2a_1a_2}{\sqrt{(a_1^2+a_2^2)(a_2^2+a_3^2)}}+\frac{2a_2a_3}{\sqrt{(a_2^2+a_3^2)(a_3^2+a_4^2)}}+\frac{2a_3a_4}{\sqrt{(a_3^2+a_4^2)(a_4^2+a_1^2)}}$$$$+\frac{2a_4a_1}{\sqrt{(a_4^2+a_1^2)(a_1^2+a_2^2)}}+\frac{2a_1a_3}{\sqrt{(a_1^2+a_3^2)(a_3^2+a_4^2)}}+\frac{2a_2a_4}{\sqrt{(a_2^2+a_3^2)(a_4^2+a_1^2)}}\leqslant 9.$$一方面, 设 $a_1^2+a_2^2$ 为 $4$ 个分母中的最小者, 则 $\frac{a_1^2}{a_1^2+a_2^2}+\frac{a_2^2}{a_2^2+a_3^2}+\frac{a_3^2}{a_3^2+a_4^2}+\frac{a_4^2}{a_4^2+a_1^2}\leqslant\frac{a_1^2}{a_1^2+a_2^2}+\frac{a_2^2}{a_1^2+a_2^2}+1+1=3$. 另一方面, 根据 Cauchy-Schwarz 不等式, $\frac{2a_1a_2}{\sqrt{(a_1^2+a_2^2)(a_2^2+a_3^2)}}+\frac{2a_3a_4}{\sqrt{(a_3^2+a_4^2)(a_4^2+a_1^2)}}\leqslant\frac{2a_1a_2}{a_1a_2+a_2a_3}+\frac{2a_3a_4}{a_3a_4+a_4a_1}=2$. 同理 $\frac{2a_2a_3}{\sqrt{(a_2^2+a_3^2)(a_3^2+a_4^2)}}+\frac{2a_4a_1}{\sqrt{(a_4^2+a_1^2)(a_1^2+a_2^2)}}\leqslant 2$, $\frac{2a_1a_3}{\sqrt{(a_1^2+a_3^2)(a_3^2+a_4^2)}}+\frac{2a_2a_4}{\sqrt{(a_2^2+a_3^2)(a_4^2+a_1^2)}}\leqslant 2$. 将其相加即得结论. 我们下面考虑一般的情况. 若 $n=4k$, 则有 $$\begin{aligned}\sum\limits_{i=1}^{4k} {\frac{{a}_{i}}{\sqrt{{a}_{i}^{2}+{a}_{{i}+{1}}^{2}+{\cdots}{{+}}{a}_{{i}{+}{k}}^{2}}}}&\leqslant\sum\limits_{i=1}^{4k}\frac{a_i}{\sqrt{a_i^2+a_{i+k}^2}}=\sum\limits_{i=1}^k\left(\frac{a_i}{\sqrt{a_i^2+a_{i+k}^2}}+\frac{a_{i+k}}{\sqrt{a_{i+k}^2+a_{i+2k}^2}}+\frac{a_{i+2k}}{\sqrt{a_{i+2k}^2+a_{i+3k}^2}}+\frac{a_{i+3k}}{\sqrt{a_{i+3k}^2+a_{i}^2}}\right)\\&\leqslant\sum\limits_{i=1}^k3=3k=n-k.\end{aligned}$$因此成立. 若 $n>4k$, 我们对 $n$ 运用数学归纳法, 当 $n=4k$ 时已经成立. 假设对 $n-1$ 成立, 考虑 $n$ 的情形. 不妨设 $a_n=\max _{1\leq i\leq n}a_i$, 则对 $n-k\leq i\le n-1$, ${\frac{{a}_{i}}{\sqrt{{a}_{i}^{2}+{a}_{{i}+{1}}^{2}+{\cdots}{{+}}{a}_{{i}{+}{k}}^{2}}}}\leqslant\frac{a_i}{\sqrt{a_i^2+\cdots +a_{n-1}^2+a_{n+1}^2+\cdots +a_{i+k+1}^2}}$. 对 $1\le i\le n-1$, 令 $b_i=a_i$, 这里 $b_i$ 脚标按照$\mod n-1$ 理解. 结合 $\frac{{a}_{n}}{\sqrt{{a}_{n}^{2}+{a}_{{n}+{1}}^{2}+{\cdots}{{+}}{a}_{{n}{+}{k}}^{2}}}<1$, 可得 $$\sum\limits_{i=1}^{n} {\frac{{a}_{i}}{\sqrt{{a}_{i}^{2}+{a}_{{i}+{1}}^{2}+{\cdots}{{+}}{a}_{{i}{+}{k}}^{2}}}}<1+\sum\limits_{i=1}^{n-1}\frac{b_i}{\sqrt{b_i^2+\cdots +b_{i+k}^2}}<1+n-k-1=1.$$因此结论成立.$\blacksquare$

note

You got this from 数之谜, right?

youthdoo wrote: You got this from 数之谜, right? Yes, you think I have any idea on this???

But... How come I know whether you have any idea? By the way, how did you manage to copy it out?