An inequality improvement on AMM 11145 I have asked the same question in math.stackexchange, I am reposting it here, looking for answers:
How to show that for $a_1,a_2,\cdots,a_n >0$ real numbers and for $n \ge 3$: 
$$\sum_{k=1}^{n}\dfrac{k}{a_{1}+a_{2}+\cdots+a_{k}}\le\left(2-\dfrac{7\ln{2}}{8\ln{n}}\right)\sum_{k=1}^{n}\dfrac{1}{a_{k}}$$

I came up with something years ago which is similar to Fedor Petrov's (or other users'). 
Problem: Let $a_i > 0; \ i = 1, 2, \cdots, n$ ($n\ge 2$). Let $C(n) = 2 - \frac{7\ln 2}{8\ln n}$. Prove that
$$\sum_{k=1}^n \frac{k}{a_1 + a_2 + \cdots + a_k}
\le C(n)\sum_{k=1}^n \frac{1}{a_k}.$$
Introducing the coefficients (to be determined) $C_1, C_2, \cdots, C_n > 0$, by the Cauchy-Bunyakovsky-Schwarz inequality, we have
$$\frac{k}{a_1 + a_2 + \cdots + a_k} \le \frac{k}{(C_1+C_2 + \cdots + C_n)^2}\sum_{i=1}^k \frac{C_i^2}{a_i}$$
and
\begin{align}
\sum_{k=1}^n \frac{k}{a_1 + a_2 + \cdots + a_k} 
&\le \sum_{k=1}^n \frac{k}{(C_1+C_2 + \cdots + C_n)^2}\sum_{i=1}^k \frac{C_i^2}{a_i}\\
&=  \sum_{k=1}^n \Big[C_k^2\sum_{m=k}^n \frac{m}{(C_1+C_2+\cdots+C_m)^2}\Big]\frac{1}{a_k}.
\end{align}
Equality occurs if and only if $C_k = a_k, \ k=1, 2, \cdots, n$.
The problem becomes: Can we choose $C_k > 0,\ k=1, 2, \cdots, n$ such that 
$$\sup_{k=1, 2, \cdots, n} C_k^2\sum_{m=k}^n \frac{m}{(C_1+C_2+\cdots+C_m)^2} \le C(n), \ \forall n\ge 2 ?$$
It is easy if $C(n)=2$. For a simple proof, I chose $$C_k = \sqrt{k(k+1)(k+2)(k+3)} - \sqrt{(k-1)k(k+1)(k+2)}, \ k = 1, 2, \cdots, n.$$
We have $C_1 + C_2 + \cdots + C_m = \sqrt{m(m+1)(m+2)(m+3)}$ and
\begin{align}
&C_k^2\sum_{m=k}^n \frac{m}{(C_1+C_2+\cdots+C_m)^2}\\
= \ &k(k+1)(k+2)\big(2k+2 - 2\sqrt{(k+3)(k-1)}\big)\\
&\quad \cdot\Big(\frac{1}{2(k+1)} - \frac{1}{2(k+2)} - \frac{1}{2(n+2)(n+3)}\Big)\\
\le \ & k(k+1)(k+2)\big(2k+2 - 2\sqrt{(k+3)(k-1)}\big)
\Big(\frac{1}{2(k+1)} - \frac{1}{2(k+2)}\Big)\\
\le \ & 2.
\end{align}
Fedor Petrov chose $C_k = k,\ \forall k$. We have
$$C_k^2\sum_{m=k}^n \frac{m}{(C_1+C_2+\cdots+C_m)^2} = 
k^2\Big(-\frac{4}{n+1}+4\Psi(1, n+2)+\frac{4}{k}-4\Psi(1, 1+k)\Big).$$
Denote the RHS as $f(k, n)$. 
We have the asymptotic expansion $f(\sqrt{n}, n) \sim 2 - \frac{2}{3\sqrt{n}} - \frac{2}{n} + \cdots (n\to \infty)$.
Thus, it at most gives $C(n) \ge 2 - \frac{A}{\sqrt{n}}$ where $A$ is a constant.
Fedor Petrov also suggested $C_k = k + \lambda$ (optimize $\lambda$). If $\lambda$ is a constant, similarly,
we have $f(\sqrt{n},n, \lambda) \sim 2 - \frac{4\lambda + 2}{3\sqrt{n}} - \cdots (n \to \infty)$.
It is not enough. 
