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

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

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

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.

• Your proof seems perfectly fine to me! Aug 15, 2022 at 3:55
• If I were writing this out for a competition or an exam, I would add some detail on how: $$\sum\limits_{j=1}^n\left(x_j\sum\limits_{i=1}^nx_i - x_j\right) = 2\sum\limits_{1\leq i < j\leq n}x_ix_j$$ Aug 15, 2022 at 4:10
• Mention that $x_j>0$ Aug 15, 2022 at 5:08
• I edited the question @peterwhy thank you.
– User
Aug 15, 2022 at 12:50
• @Youem Thank you for checking and your time.
– User
Aug 15, 2022 at 14:22

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 $$f(x_j)=\frac{x_j}{s-x_j} \implies f''(x)=\frac{2s}{(s-x_j)^3}>0,\quad 0 So by Jensen's inequality $$\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}.$$