Local Mathematical Competitions

We claim $c_{\min }=2$. Taking $a_j=\frac{1}{j}$ for $j=1,2,\dots ,n$, we have $$\sum _{k=1}^n\frac{k}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_k}}=\sum _{k=1}^n\frac{k}{1+2+\cdots +k}=2\sum _{k=1}^n\frac{1}{k+1},\text{ while }c\sum _{k=1}^n{a}_k=$$

We will now prove that $c=2$ is suitable. From the Cauchy-Schwartz inequality, $$\begin{gathered} \frac{k^2(k+1{)}^2}{4}={\left(\sum _{j=1}^{k}j\right)}^2\le \left(\sum _{j=1}^k{j}^2{a}_j\right)\left(\sum _{j=1}^k\frac{1}{a_j}\right),\text{ hence} \\ \frac{k}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_k}}\le \frac{4}{k(k+1{)}^2}\sum _{j=1}^k{j}^2{a}_j. \end{gathered}$$ Therefore $$\begin{array}{rl}\sum _{k=1}^n\frac{k}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_k}}& \le \sum _{k=1}^n\left(\frac{4}{k(k+1{)}^2}\sum _{j=1}^k{j}^2{a}_j\right)=\\ & =\sum _{j=1}^n\left(j^2{a}_j\sum _{k=j}^n\frac{4}{k(k+1{)}^2}\right)=2\sum _{j=1}^n\left(j^2{a}_j\sum _{k=j}^n\frac{2k}{k^2(k+1{)}^2}\right)<\\ & <2\sum _{j=1}^n\left(j^2{a}_j\sum _{k=j}^n\frac{2k+1}{k^2(k+1{)}^2}\right).\text{ But}\\ \sum _{k=j}^n\frac{2k+1}{k^2(k+1{)}^2}& =\sum _{k=j}^n\left(\frac{1}{k^2}-\frac{1}{(k+1{)}^2}\right)=\frac{1}{j^2}-\frac{1}{(n+1{)}^2}<\frac{1}{j^2},\\ \text{hence }\sum _{k=1}^n\frac{k}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_k}}& <2\sum _{k=1}^n{a}_k.\end{array}$$