MT【140】是否存在常数$\textbf{C}$

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(中国第59届国际数学奥林匹克国家集训队2018.3.20日测试题)
证明:存在常数$C>0$使得对于任意的正整数$m$,以及任意$m$个正整数$a_1,a_2,\cdots,a_m$,都有
$H(a_1)+H(a_2)+\cdots+H(a_m)\le C\left(\sum\limits_{k=1}^m{ka_k}\right)^{\frac{1}{2}}$,其中$H(n)=\sum\limits_{k=1}^{n}{\dfrac{1}{k}}$

证明:存在.$C=2$满足要求.记$\{a_1,a_2,\cdots,a_m\}=\{b_1,b_2\cdots,b_m\}$其中$b_1\ge b_2\ge \cdots \ge b_m$

$$\begin{align*} LHS&=\dfrac{1}{1}+\dfrac{1}{2}+\cdots+\dfrac{1}{b_1}+\dfrac{1}{1}+\dfrac{1}{2}+\cdots+\dfrac{1}{b_2} +\dfrac{1}{1}+\dfrac{1}{2}+\cdots+\dfrac{1}{b_m}\\ & \le m\left(\dfrac{1}{1}+\dfrac{1}{2}+\cdots+\dfrac{1}{b_m}\right) \\ & \le m\sqrt{(1^2+1^2+\cdots 1^2)(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\cdots+\dfrac{1}{b_m^2})}\quad(\textbf{此处用到柯西不等式})\\ &\le m\sqrt{b_m\cdot(1+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}\cdots+\dfrac{1}{b_m-1}-\dfrac{1}{b_m})} \quad (\textbf{此处用到}\dfrac{1}{k^2}\le\dfrac{1}{k-1}-\dfrac{1}{k})\\ &=m\sqrt{2b_m-1}\\ RHS&=C\sqrt{1a_1+2a_2+\cdots+ma_m}\\ &\ge C\sqrt{(1+2+\cdots m)b_m}\\ &=C\sqrt{\dfrac{m(m+1)}{2}b_m} \end{align*}$$

取$C=2$时 $2\sqrt{\dfrac{m(m+1)}{2}b_m}\ge m\sqrt{2b_m-1}$显然成立.