<s>[INST] Inequality for harmonic means Prove that for real numbers $a_1 ,a_2 ,...,a_n >0$ the following inequality holds
$$\frac{1}{a_1 } +\frac{2}{a_1 +a_2 } +...+\frac{n}{a_1 +a_2 +...+a_n }\leq 4\cdot \left(\frac{1}{a_1} +\frac{1}{a_2 } +...+\frac{1}{a_n} \right).$$ 
 [/INST] Upper Bound
By Cauchy-Schwarz, we have
$$
\begin{align}
\left(\sum_{j=1}^ka_j\right)\left(\sum_{j=1}^k\frac{j^2}{a_j}\right)
&\ge\left(\sum_{j=1}^kj\right)^2\\[3pt]
&=\frac{k^2(k+1)^2}{4}\tag{1}
\end{align}
$$
Thus,
$$
\begin{align}
\sum_{k=1}^n\frac{k}{\sum\limits_{j=1}^ka_j}
&\le\sum_{k=1}^n\frac4{k(k+1)^2}\sum_{j=1}^k\frac{j^2}{a_j}\\
&=\sum_{j=1}^n\frac{j^2}{a_j}\sum_{k=j}^n\frac4{k(k+1)^2}\\
&\le\sum_{j=1}^n\frac{j^2}{a_j}\sum_{k=j}^n2\left(\frac1{k^2}-\frac1{(k+1)^2}\right)\\
&\le\sum_{j=1}^n\frac{j^2}{a_j}\frac2{j^2}\\
&=2\sum_{j=1}^n\frac1{a_j}\tag{2}
\end{align}
$$
Thus, the ratio is at most $2$.

Sharpness
Set $a_k=k^\beta$ for $0\lt\beta\lt1$. First
$$
\begin{align}
\sum_{k=1}^n\frac{k}{\sum\limits_{j=1}^ka_j}
&=\sum_{k=1}^n\frac{k}{\frac1{1+\beta}k^{\beta+1}+O(k^\beta)}\\
&=(1+\beta)\sum_{k=1}^n\frac1{k^\beta+O(k^{\beta-1})}\\
&=\frac{1+\beta}{1-\beta}n^{1-\beta}+O(1)\\[3pt]
&\sim\frac{1+\beta}{1-\beta}n^{1-\beta}\tag{3}
\end{align}
$$
Next
$$
\begin{align}
\sum_{k=1}^n\frac1{a_k}
&=\sum_{k=1}^n\frac1{k^\beta}\\
&=\frac1{1-\beta}n^{1-\beta}+O(1)\\[3pt]
&\sim\frac1{1-\beta}n^{1-\beta}\tag{4}
\end{align}
$$
Thus, as $\beta\to1^-$, the ratio tends to $2$ for large $n$. Therefore, $2$ is sharp.
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