MT【140】是否存在常数$\textbf{C}$

\(\textbf{天下事有难易乎?为之,则难者亦易矣 不为,则易者亦难矣------《为学》}\)

(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.