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suppose $x_i>0$ for $i=1,2,\cdots,n$ and $x_1+x_2+\cdots+x_n=1$ show that $$\frac{x_1}{1+x_2+x_3+\cdots+x_n}+\frac{x_2}{1+x_1+x_3+\cdots+x_n}+\frac{x_3}{1+x_1+x_2+\cdots+x_n}+\cdots+\frac{x_n}{1+x_1+x_2+\cdots+x_{n-1}}\ge\frac{n}{2n-1}$$ attempt

from the equalitys i could rewrite $1-x_1=x_2+\cdots+x_n$ and then $$\frac{x_1}{1+x_2+x_3+\cdots+x_n}+\frac{x_2}{1+x_1+x_3+\cdots+x_n}+\frac{x_3}{1+x_1+x_2+\cdots+x_n}+\cdots+\frac{x_n}{1+x_1+x_2+\cdots+x_{n-1}}=\frac{x_1}{2-x_1}+\frac{x_2}{2-x_2}+\frac{x_3}{2-x_3}+\cdots+\frac{x_n}{2-x_n}$$ but idk how to proced from there, so how do i solve this? the fact $x_i>0$ makes me assume AM-GM inequality could be used but idk how one would use.

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Use Jensen's inequality.

Since $x_i\gt 0$ and $\sum_i x_i=1$, we also have $x_i\lt 1~\forall~i$ and the map $f\colon x\mapsto\dfrac x{2-x}$ is convex on $[0,1]$

We have,

$$\sum_{\text{cyc}}\frac{x_i}{2-x_i}=\sum_{\text{cyc}}f(x_i)\geq nf\left(\frac{\sum_i x_i}n\right)=nf(1/n)=n\frac{1/n}{2-1/n}=\frac n{2n-1}$$

with equality iff all the $x_i$'s are equal.


Alternatively, note that $\dfrac{x_i}{2-x_i}=\dfrac 1{\frac 2{x_i}-1}$ for $x_i\ne 0$; since $0\lt x_i\lt 1$ as well, we have $c_i=\dfrac 2{x_i}\gt 2$, or $c_i-1\gt 1$, or $0\lt\dfrac 1{c_i-1}\lt 1$

Now, use the HM-AM inequality on $\{c_i-1\}_i$ (more generally, the power mean inequality $M_{-1}\le M_1$).

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    $\begingroup$ This has been asked and answered before. I suggest to use Approach0 to search for duplicates, before answering a question. See also How to search on this site? $\endgroup$ – Martin R Jun 21 at 7:40
  • $\begingroup$ @MartinR: Thank you, indeed searching for a duplicate first would have been a better idea. I knew about Approach0 but I admit I didn't put effort into searching. Glad to see quality standards finally being enforced! :) Will do from next time. +1 $\endgroup$ – Prasun Biswas Jun 21 at 14:42
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By C-S we obtain: $$\sum_{i=1}^n\frac{x_i}{1+\sum\limits_{k\neq i}x_k}=-n+\sum_{i=1}^n\left(\frac{x_i}{1+\sum\limits_{k\neq i}x_k}+1\right)=-n+\sum_{i=1}^n\frac{2}{1+\sum\limits_{k\neq i}x_k}=$$ $$=-n+\frac{2}{2n-1}\sum_{i=1}^n\left(1+\sum\limits_{k\neq i}x_k\right)\sum_{i=1}^n\frac{1}{1+\sum\limits_{k\neq i}x_k}\geq-n+\frac{2n^2}{2n-1}=\frac{n}{2n-1}.$$

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    $\begingroup$ welcome back sir! $\endgroup$ – Albus Dumbledore Jun 21 at 3:39
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    $\begingroup$ It took me less than a minute to find an identical question with answers (using Approach0). How much effort did you put into searching for a duplicate? $\endgroup$ – Martin R Jun 21 at 7:38
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After OP's attempt, let $$F=\frac{x_1}{2-x_1}+\frac{x_2}{2-x_2}+\frac{x_3}{2-x_3}+\cdots+\frac{x_n}{2-x_n}.$$ Let $x_1\ge x_2\ge x_3.....\ge x_n$, apply Tchebecheff's inequality that if $a_1\ge a_2 \ge a_3,....\ge a_n$ and $b_1 \ge b_2 \ge b_3 \ge ...\ge b_n$, then $$\sum_{k=1}^{n} a_kb_k \ge \frac{1}{n}\sum_{k=1}^{n} a_k \sum_{k=1}^{n} b_k$$ Here let $x_1\ge x_2 \ge x_3,....\ge x_n$ then $(2-x_1)^{-1} \ge (2-x_2)^{-1} \ge (2-x_3)^{-1} \ge ...\ge (2-x_n)^{-1}$, so we write $$F\ge \frac{1}{n}\sum_{k=1}^{n} x_k \sum_{k=1}^{n} \frac{1}{2-x_k}$$ Next by AM-HM $\sum_k a_k \sum_k \frac{1}{a_k} \ge n^2$, we get $$F\ge \frac{1}{n} \sum_{k=1}^n x_k ~~ \frac{n^2}{2n-\sum_k x_k}=\frac{n}{2n-1}.$$

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    $\begingroup$ It took me less than a minute to find an identical question with answers (using Approach0). How much effort did you put into searching for a duplicate? $\endgroup$ – Martin R Jun 21 at 7:38
  • $\begingroup$ @Martin R To find whether a problem is duplicate I mock as if I am asking/posting this problem then some times only REVIEW your question works. also It is not guaranteed if it is there. Do you know any other method please suggest. There should be a "search" section where we can find things of our interes in MSE. Fior instance I want to search results having Frullani integral. Please suggest how can I find it. $\endgroup$ – Z Ahmed Jun 21 at 7:47
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    $\begingroup$ See How to search on this site?. – Approach0 and SearchOnMath work very well. $\endgroup$ – Martin R Jun 21 at 7:50
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    $\begingroup$ @ZAhmed Google search "frullani integral site:math.stackexchange.com". Search once, search harder, search even harder, then answer. Ideally more time should be spent searching and redirecting, than answering. In fact, the only reason why searching is so difficult is because of rampant duplication in the first place, which creates many copies of the same concept and then people don't know where to search for what. $\endgroup$ – Teresa Lisbon Jun 21 at 11:55
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    $\begingroup$ What happened to the civility? $\endgroup$ – Toby Mak Jun 28 at 13:08

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