This general inequality maybe is true? $\sum_{i=1}^{n}\frac{i}{1+a_{1}+\cdots+a_{i}}<\frac{n}{2}\sqrt{\sum_{i=1}^{n}\frac{1}{a_{i}}}$

Question

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

Related: http://math.stackexchange.com/questions/214634/prove-that-sum-k-1n-frac2k1a-1a-2-a-k4-sum-k-1n-frac1a-k — Jack D'Aurizio, Jun 30 '15 at 15:12
I know, but maybe they can be tackled in a similar, variational-style, way. — Jack D'Aurizio, Jun 30 '15 at 15:25
I was about to note that in your second equation, in the numerator on the left side, you probably wanted an $i$. However, I see that sranthrop has posted a counterexample. I don't know if it helps with your problem, but this answer seems related. — robjohn, Jul 01 '15 at 14:24
@robjohn,first,Thank you for your comment, I have try this methods,can't solve it.maybe you can — math110, Jul 01 '15 at 14:28

chenbai · Accepted Answer · 2015-07-02T00:36:26.683

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

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? — A. Thomas Yerger, Jul 01 '15 at 23:20
Oh,It's very nice,I forgot this inequality can use mathematical induction; — math110, Jul 02 '15 at 05:15

score 0 · Answer 2 · edited Apr 13 '17 at 12:21

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*}

I know it is not exactly an answer to the question, but maybe it helps. — sranthrop, Jul 01 '15 at 17:33

This general inequality maybe is true? $\sum_{i=1}^{n}\frac{i}{1+a_{1}+\cdots+a_{i}}<\frac{n}{2}\sqrt{\sum_{i=1}^{n}\frac{1}{a_{i}}}$

2 Answers2