Suppose $x_i>0$ for $i=1,2,\cdots,n$ and $x_1+x_2+\cdots+x_n=1 \ldots$ [duplicate]

Question

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.

Answer 1

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.

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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 $M_{-1}\le M_1$).

Answer 2

By C-S we obtain: $$\sum_{i=1}^n\frac{x_i}{1+\sum\limits_{k\neq i}x_k}=-n+\sum_{i=1}^n\left(\frac{x_i}{1+\sum\limits_{k\neq i}x_k}+1\right)=-n+\sum_{i=1}^n\frac{2}{1+\sum\limits_{k\neq i}x_k}=$$ $$=-n+\frac{2}{2n-1}\sum_{i=1}^n\left(1+\sum\limits_{k\neq i}x_k\right)\sum_{i=1}^n\frac{1}{1+\sum\limits_{k\neq i}x_k}\geq-n+\frac{2n^2}{2n-1}=\frac{n}{2n-1}.$$

Answer 3

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 $$F\ge \frac{1}{n} \sum_{k=1}^n x_k ~~ \frac{n^2}{2n-\sum_k x_k}=\frac{n}{2n-1}.$$