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?


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

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

Just an idea :

We try the substitution :


We have :


Wich have the form :


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 !


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.