Eighteenth STARS OF MATHEMATICS Competition

Let $A_j=\{i:2^{j-1}\le a_i<2^j\},j=1,2,\dots ,11$. The $A_j$ form a partition of the index set $\{1,2,\dots ,n\}$.

Note that, for every $j$ in the range 1 through 11, if $A_j$ is non-empty, then $$\begin{array}{rl}\sum _{i\in A_j}\frac{1}{a_i}(a_1+a_2+\cdots +a_i)& =\sum _{i\in A_j}\left(\frac{a_1+a_2+\cdots +a_{i-1}}{a_i}+1\right)\\ & >\sum _{k=0}^{|A_j|-1}\frac{k\cdot 2^{j-1}}{2^j}+|A_j|=\frac{1}{2}\sum _{k=0}^{|A_j|-1}k+|A_j|\\ & =\frac{1}{4}|A_j|(|A_j|-1)+|A_j|=\frac{1}{4}|A_j|(|A_j|+3);\end{array}$$

As the $A_j$ form a partition of the index set $\{1,2,\dots ,n\}$ and at least one $A_j$ is non-empty, $$ $$\begin{array}{rl}\sum _{i=1}^n\frac{1}{a_i}(a_1+a_2+\cdots +a_i)& =\sum _{j=1}^{11}\sum _{i\in A_j}\frac{1}{a_i}(a_1+a_2+\cdots +a_i)\\ & >\frac{1}{4}\sum _{j=1}^{11}|A_j|(|A_j|+3)=\frac{1}{4}\left(\sum _{j=1}^{11}|A_j{|}^2+3n\right)\\ & \ge \frac{1}{4}\left(\frac{n^2}{11}+3n\right)=\frac{1}{44}n(n+33),\text{ by Cauchy-Schwarz.}\end{array}$$