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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

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  • $\begingroup$ Related: math.stackexchange.com/questions/214634/… $\endgroup$ – Jack D'Aurizio Jun 30 '15 at 15:12
  • $\begingroup$ Hello,@jack,there are different problem $\endgroup$ – math110 Jun 30 '15 at 15:13
  • $\begingroup$ I know, but maybe they can be tackled in a similar, variational-style, way. $\endgroup$ – Jack D'Aurizio Jun 30 '15 at 15:25
  • $\begingroup$ Numerical tests show that (2) is wrong for $n=3$. $\endgroup$ – sranthrop Jun 30 '15 at 15:28
  • $\begingroup$ @sranthrop,oh, can you post your counter-example? Thank you $\endgroup$ – math110 Jun 30 '15 at 15:30
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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

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  • $\begingroup$ I want to read this, but it's just so cumbersome. Can you format it so that the lines are a little further spaced out? $\endgroup$ – Alfred Yerger Jul 1 '15 at 23:20
  • $\begingroup$ so easy... nice! +1 $\endgroup$ – sranthrop Jul 2 '15 at 0:40
  • $\begingroup$ Oh,It's very nice,I forgot this inequality can use mathematical induction; $\endgroup$ – math110 Jul 2 '15 at 5:15
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    $\begingroup$ at last also can use Cauchy-Schwarz inequality $\endgroup$ – math110 Jul 5 '15 at 4:34
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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*}

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  • $\begingroup$ I know it is not exactly an answer to the question, but maybe it helps. $\endgroup$ – sranthrop Jul 1 '15 at 17:33

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