Prove that: $\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\ge\frac{n}{n-1}$

Question

$\color{red}{\textbf{Problem:}}$

Let, $x_i>0,1\le i\le n$, then

Prove that: $$\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\ge\frac{n}{n-1}$$

$\color{red}{\textbf{Proof:}}$

Using AM-GM inequality we have,

\begin{align}&\sum_{1\le i<j\le n} (x_i^2+x_j^2)=(n-1)\sum_{1\le i\le n} x_i^2\ge 2\left(\sum_{1\le i < j \le n}x_ix_j\right)\\ \implies &\sum_{1\le i\le n} x_i^2\ge \frac 2{n-1}\left(\sum_{1\le i < j \le n}x_ix_j\right)\\ \implies &\left(\sum_{1\le i\le n} x_i\right)^2\ge\left( 2+\frac 2{n-1}\right)\left(\sum_{1\le i < j \le n}x_ix_j\right)\\ &\qquad\qquad\quad =\frac{2n}{n-1}\left(\sum_{1\le i < j \le n}x_ix_j\right)\end{align}

Finally, applying Cauchy-Schwars, we obtain

\begin{align}\left(\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\right)\left(\sum_{1\le j\le n} \left({x_j}{\sum_{1\le i\le n} x_i-x_j}\right)\right)&\ge \frac{ \left(\sum_{1\le j\le n} x_i\right)^2}{2\left(\sum_{1\le i < j \le n}x_ix_j\right)}\\ &\ge \frac {\frac{2n}{n-1}\sum_{1\le i < j \le n}x_ix_j}{2\sum_{1\le i < j \le n}x_ix_j}\\ &=\boxed {\frac{n}{n-1}.}\end{align}

I need to know if there is something wrong with my solution.

Answer 1

OP's work is correct however $x_j>0$ needs to be mentioned. Here is another approach:

Let $s=\sum_{j=1}^{n} x_j,$ then $$f(x_j)=\frac{x_j}{s-x_j} \implies f''(x)=\frac{2s}{(s-x_j)^3}>0,\quad 0<x_j<s.$$ So by Jensen's inequality $$\frac{1}{n}\sum_{j=1}^{n} f(x_j) \ge f(\frac{s}{n}) \implies \sum_{j=1}^{n} \frac{x_j}{s-x_j}\ge n \frac{s/n}{s-s/n}=\frac{n}{n-1}. $$