# Prove a strong inequality $\sum_{k=1}^n\frac{k}{a_1+a_2+\cdots+a_k}\le\left(2-\frac{7\ln 2}{8\ln n}\right)\sum_{k=1}^n\frac 1{a_k}$

For $$a_i>0$$ ($$i=1,2,\dots,n$$), $$n\ge 3$$, prove that $$\sum_{k=1}^n\frac{k}{a_1+a_2+\cdots+a_k}\le\left(2\color{red}{-\frac{7\ln 2}{8\ln n}}\right)\sum_{k=1}^n\frac 1{a_k}.$$

The case without $$\color{red}{-\dfrac{7\ln 2}{8\ln n}}$$ could be shown here. I have no idea how the $$\color{red}{\text{red}}$$ term comes from.

Note: This question should not be closed although there was a duplicated question $$4$$ years ago (see here). Duplicate of unanswered question suggests that if there is no accepted answer in the old question, the new question can stay open in the hope of attracting an answer.

The question comes from the Chinese Mathematical Olympiad training team and there is no answer provided.

# Source:

• See Q.25 here (one of the official accounts that provides Chinese MO questions on January $$23^{\rm rd}$$, $$2018$$)
• It has also appeared here (A blog from the person who set this question on December $$17^{\rm th}$$, $$2013$$).
• This looks like a problem you have collected from / inspired by some source. According to recent discussions in Meta, we are looking forward to including sources for all applicable questions. Can you provide the source by editing the question?Refer-math.meta.stackexchange.com/questions/29290/… – tatan Oct 22 '18 at 14:10
• See question #25 here. The question first appeared here from the same person. – Tianlalu Oct 22 '18 at 14:42