# Sum inequality $x_1, x_2, …, x_n \gt 0 \in \mathbb{R}$ and $\sum_{i=1}^n \frac{x_i}{s-x_i+1}\le 1$

Be $n \ge 2 \in \mathbb{N}$ and $x_1, x_2, ..., x_n \gt 0 \in \mathbb{R}$, with $s=x_1+x_2+..+x_n$ and: $$\sum_{i=1}^n \frac{x_i}{s-x_i+1}\le 1$$

Prove that:

$$\sum_{i=1}^n \frac{1}{s-x_i+1}\ge \frac{1}{n-1}$$

I have no idea on how to do this, or if it is really true. Can you help me please? Thank you very much!

## 1 Answer

Put $y_i = s - x_i+1>0\implies x_i = s-y_i+1\implies \displaystyle \sum_{i=1}^n\dfrac{s-y_i+1}{y_i}\le 1\implies (s+1)\displaystyle \sum_{i=1}^n\dfrac{1}{y_i}-n\le 1\implies \displaystyle \sum_{i=1}^n\dfrac{1}{y_i}\le \dfrac{n+1}{s+1}$. From an application of the AM-GM inequality twice: $\left((n-1)s+n\right)\dfrac{n+1}{s+1} =\dfrac{n+1}{s+1}\displaystyle \sum_{i=1}^n y_i\ge \displaystyle \sum_{i=1}^n y_i\displaystyle \sum_{i=1}^n \dfrac{1}{y_i}\ge n^2\implies (n^2-1)s+n(n+1)\ge n^2(s+1)\implies s \le n\implies \displaystyle \sum_{i=1}^n \dfrac{1}{y_i} \ge \dfrac{n^2}\sum_{i=1}^n y_i}=\dfrac{n^2}{(n-1)s+n}\ge \dfrac{n^2}{(n-1)n+n}=1 \ge \dfrac{1}{n-1$ since $n \ge 2$. We're done.