This general inequality maybe is true? $\sum_{i=1}^{n}\frac{i}{1+a_{1}+\cdots+a_{i}}<\frac{n}{2}\sqrt{\sum_{i=1}^{n}\frac{1}{a_{i}}}$ 
Let $a_{1},a_{2},\ldots,a_{n}>0$ and prove or disprove
  $$\dfrac{1}{1+a_{1}}+\dfrac{2}{1+a_{1}+a_{2}}+\cdots+\dfrac{n}{1+a_{1}+a_{2}+\cdots+a_{n}}\le\dfrac{n}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}}\tag{1}$$

This problem from when I solve this two variable inequality
since $n=1$ it is clear
$$\dfrac{1}{1+a_{1}}\le\dfrac{1}{2}\sqrt{\dfrac{1}{a_{1}}}$$
because $1+a_{1}\ge 2\sqrt{a_{1}}$
$n=2$ case,can see this links my answer.
For general 
simaler this two variable inequality methods, then I  use Cauchy-Schwarz inequality we have
$$\left(\sum_{i=1}^{n}\dfrac{1}{1+a_{1}+\cdots+a_{i}}\right)^2\le\left(\sum_{i=1}^{n}\dfrac{1}{a_{i}}\right)\cdot\left(\sum_{i=1}^{n}\dfrac{i^2a_{i}}{(1+a_{1}+a_{2}+\cdots+a_{i})^2}\right)$$
it suffices to show that
$$\sum_{i=1}^{n}\dfrac{i^2a_{i}}{(1+a_{1}+\cdots+a_{i})^2}\le\dfrac{n^2}{4}\tag{2}$$
it seem hard.
because I tried following also  fail;
$$\sum_{i=1}^{n}\dfrac{i^2a_{i}}{(1+a_{1}+\cdots+a_{i})^2}<\sum_{i=1}^{n}i^2\left(\dfrac{1}{1+a_{1}+\cdots+a_{i-1}}-\dfrac{1}{1+a_{1}+a_{2}+\cdots+a_{i}}\right)$$
and use   Abel transformation.not can  to prove $(2)$,
Note $(1)$ Left side hand  was simaler Hardy's inequality when $p=-1$,But there are different problem.
EDIT:Numerical tests $(2)$ is not right.so my idea can't works
 A: suppose $n=k, \\ \dfrac{1}{1+a_{1}}+\dfrac{2}{1+a_{1}+a_{2}}+\cdots+\dfrac{k}{1+a_{1}+a_{2}+\cdots+a_{k}} \\ \le\dfrac{k}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{k}}}$
when $n=k+1 $
LHS=$\dfrac{1}{1+a_{1}}+\dfrac{2}{1+a_{1}+a_{2}}+\cdots+\dfrac{k}{1+a_{1}+a_{2}+\cdots+a_{k}}+\dfrac{k+1}{1+a_{1}+a_{2}+\cdots+a_{k+1}}  \\<\dfrac{k}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{k}}}+\dfrac{k+1}{1+a_{1}+a_{2}+\cdots+a_{k+1}} \\ <\dfrac{k}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{k}}+\dfrac{1}{a_{k+1}}}+\dfrac{k+1}{1+a_{1}+a_{2}+\cdots+a_{k+1}}$
RHS$=\dfrac{k}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{k}}+\dfrac{1}{a_{k+1}}}+\dfrac{1}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{k}}+\dfrac{1}{a_{k+1}}}$
so is remains:
$ \dfrac{k+1}{1+a_{1}+a_{2}+\cdots+a_{k+1}} \\ \le \dfrac{1}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{k}}+\dfrac{1}{a_{k+1}}}$
$1+a_{1}+a_{2}+\cdots+a_{k+1} \ge 2\sqrt{a_{1}+a_{2}+\cdots+a_{k+1}} \\ \implies \dfrac{k+1}{1+a_{1}+a_{2}+\cdots+a_{k+1}} \le \dfrac{k+1}{2\sqrt{a_{1}+a_{2}+\cdots+a_{k+1}}} \le \dfrac{1}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{k}}+\dfrac{1}{a_{k+1}}} \\ \iff \dfrac{(k+1)^2}{\sum_{i=1}^{k+1} a_i} \le \sum_{i=1}^{k+1} \dfrac{1}{a_i}$
the last one is true ,$HM\le AM$
QED
A: ONLY AN IDEA (and no solution, or at least a partial solution):
From this question (thx to the comments) we have
\begin{align*}
\sum_{k=1}^n\frac{k}{1+a_1+\ldots+a_k}\leq
\sum_{k=1}^n\frac{k}{(1/n+a_1)+\ldots+(1/n+a_k)}\leq2\left(\frac{1}{1/n+a_1}+\ldots+\frac{1}{1/n+a_n}\right)=2x,
\end{align*}
where $x:=\sum_{k=1}^n\frac{1}{1/n+a_k}$. Since
\begin{align*}
\sum_{k=1}^n\frac{1}{1/n+a_k}\leq\sum_{k=1}^n\frac{1}{1/n}=n^2,
\end{align*}
we have $x/n^2\leq 1$ and thus $x/n^2\leq\sqrt{x/n^2}$. Consequently, $x\leq n\sqrt{x}$, and this implies
\begin{align*}
\sum_{k=1}^n\frac{k}{1+a_1+\ldots+a_k}\leq2n\cdot\sqrt{\sum_{k=1}^n\frac{1}{1/n+a_k}}\leq2n\cdot\sqrt{\sum_{k=1}^n\frac{1}{a_k}}.
\end{align*}
