<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] The linked proof of Carleman's Inequality (in the comment) indicates the method of balancing coefficients in weighted mean inequalities. In the same spirit we can show, $$\frac{1}{a_1 } +\frac{2}{a_1 +a_2 } +...+\frac{n}{a_1 +a_2 +...+a_n } < 2 \left(\frac{1}{a_1} +\frac{1}{a_2 } +...+\frac{1}{a_n} \right)$$ for positive $a_i$'s.
We choose a set of positive real numbers $x_1,x_2,\ldots,x_n$ such that, by Cauchy-Schwarz Inequality: $$(a_1+a_2+\cdots+a_k)\left(\dfrac{x_1^2}{a_1}+\dfrac{x_2^2}{a_2}++\cdots\dfrac{x_k^2}{a_k}\right) \ge (x_1+x_2+\cdots+x_k)^2$$
$$\implies \dfrac{k}{a_1+a_2+\cdots+a_k} \le \dfrac{k}{(x_1+x_2+\cdots+x_k)^2}\left(\dfrac{x_1^2}{a_1}+\dfrac{x_2^2}{a_2}++\cdots\dfrac{x_k^2}{a_k}\right)$$
for each $k=1,2,\cdots,n$.
Adding them up from $k=1$ to $n$, we get:
$$\frac{1}{a_1 } +\frac{2}{a_1 +a_2 } +...+\frac{n}{a_1 +a_2 +...+a_n } \le \dfrac{c_1}{a_1} + \frac{c_2}{a_2}+\ldots+\frac{c_n}{a_n}$$
Where, $c_k = \dfrac{kx_k^2}{(x_1+x_2+\cdots+x_k)^2} + \dfrac{(k+1)x_k^2}{(x_1+x_2+\cdots+x_{k+1})^2}+\ldots+\dfrac{nx_k^2}{(x_1+x_2+\cdots+x_n)^2}$ for each $k=1,2,\ldots,n$.
It remains to choose a set of $\{x_k\}_{k=1}^n$ such that $\max\limits_{k \in \{1,2,\cdots,n\}}\{c_k\}$ is minimized.
For example if we plug in $x_k = k$, we have,
$c_k = k^2\left(\sum\limits_{j=k}^n \dfrac{j}{(1+2+\cdots+j)^2}\right) = 4k^2\left(\sum\limits_{j=k}^n \dfrac{1}{j(j+1)^2}\right) $
$$\le 2k^2\left(\sum\limits_{j=k}^n \dfrac{2j+1}{j^2(j+1)^2}\right) = 2k^2\left(\sum\limits_{j=k}^n \dfrac{1}{j^2} - \sum\limits_{j=k}^n \dfrac{1}{(j+1)^2}\right)$$
$$ = 2k^2\left(\dfrac{1}{k^2} - \dfrac{1}{(n+1)^2}\right) < 2$$
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