An Inequality Involving $n$ positive real numbers and their sum

Question

Consider the following problem from Problems from the Book proposed by Gabriel Dospinescu

Let $a_1, a_2, \dots, a_n$ be positive real numbers and let $S = a_1 + a_2 + \dots + a_n$ 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-Schwarz without any luck (the inequality would become too weak). Any suggestions or solutions?

Answer 1

Hints:
1. Use Karamata’s inequality for the convex function $t\mapsto \dfrac1t$. You will then need to show $$(\underbrace{na_1, na_2, \dots na_n}_{n \text{ terms}}, \underbrace{S, S, \dots, S}_{n(n-2) \text{ times}}) \succ (\underbrace{S+a_1-a_2, S+a_1-a_3, \dots S+a_n-a_{n-1}}_{n(n-1) \text{ terms}})$$

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

Answer 2

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 $\varphi(x)$ a convex function, $f(x)$ a non-negative Lebesgue-integrable function and $a\neq b\neq 0$.

I hope for you !