An Inequality Involving $n$ positive real numbers and their sum Consider the following problem from Problems from the Book proposed by Gabriel Dospinescu


  Let $a_1, a_2, ..., a_n$ be positive real numbers and let $S = a_1 + a_2 + ... + a_n$ be their sum. Prove that 
  $$ \frac{1}{n} \sum_{i = 1}^n \frac{1}{a_i} + \frac{n(n-2)}{S} \geq \sum_{i \neq j} \frac{1}{S + a_i - a_j}$$


I've been trying to solve this using standard methods such as fudging and Cauchy-Schwartz without any luck (the inequality would become too weak). Any suggestions or solutions?
Just an idea :
We try the substitution :
$$\frac{1}{a_i}=x_i$$
We have :
$$\frac{1}{Sx_ix_j+x_i-x_j}\int_{x_j}^{Sx_ix_j+x_i}\frac{x_ix_j}{Sx_ix_j+x_i-x_j}dy=\frac{x_ix_j}{Sx_ix_j+x_i-x_j}$$
Wich have the form :
$$\frac{1}{a-b}\int_{b}^{a}f(x)dx$$
So there is many possibilities and by example we can use Jensen's inequality for integrals  wich states :
$${\displaystyle \varphi \left({\frac {1}{b-a}}\int _{a}^{b}f(x)\,dx\right)\leq {\frac {1}{b-a}}\int _{a}^{b}\varphi (f(x))\,dx.}$$
For $f(x)\neq \infty$  a convex function and $a\neq b\neq 0$.
I hope for you !
Hints:
1. Use Karamata’s inequality for the convex function $t\mapsto \dfrac1t$. You will then need to show $$(\underbrace{na_1, na_2, \dots na_n}_{n \text{ terms}}, \underbrace{S, S, \dots, S}_{n(n-2) \text{ times}}) \succ (\underbrace{S+a_1-a_2, S+a_1-a_3, \dots S+a_n-a_{n-1}}_{n(n-1) \text{ terms}})$$


The majorisation can be established by noting WLOG $a_k$ can be ordered non-decreasing say...