A Generalization of an Inequality from a Romanian Olympiad

On an earlier page we discussed a problem from a Romanian olympiad and suggested its obvious generalization. Leo Giugiuc and Dan Sitaru proved a more involved generalization:

a generalization of an inequality from a Romanian Olympiad


By MacLaurin's inequality,

$\displaystyle\frac{2}{n(n-1)}\sum_{1\le i\lt j\le n}a_{i}a_{j}\le\left(\frac{1}{n}\sum_{k=1}^{n}a^k\right)^2.$

On the other hand, by Jensen's inequality,


It follows that $\displaystyle\frac{2}{n-1}\sum_{1\le i\lt j\le n}a_{i}a_{j}\le\sum_{k=1}^{n}a_k^2.\;$ Thus, suffice it to prove that


We'll use the following


Consider the function $f(x)=e^{x-1}+x\ln x-x^2,\;$ $f:\;(0,\infty )\rightarrow\,\mathbb{R}.$

Then, for every $x\gt 0,\;$ $f(x)\ge 0.$

Proof of Lemma

$f'(x)=e^{x-1}+\ln x+1-2x\;$ and $\displaystyle f''(x)=e^{x-1}+\frac{1}{x}-2.\;$ By the AM-GM inequality, $\displaystyle e^{x-1}+\frac{1}{x}\ge 2\sqrt{\frac{e^{x-1}}{x}},\;$ which, combined with the well known fact that $e^{x-1}\ge (x-1)+1=x,\;$ for $x\gt 0,\;$ gives $\displaystyle e^{x-1}+\frac{1}{x}\ge 2,\;$ and implying that, for $x\gt 0,\;$ $f''(x)\ge 0,\;$ making $f'(x)\;$ increasing.

Since $f'(x)=0,\;$ we deduce that $f(x)\;$ is decreasing on $(0,1]\;$ and increasing on $[1,\infty )$ such that it achieves its minimum at $x=1.\;$ Therefore, $f(x)\ge f(1)=0.$


Back to the main problem, we apply the lemma for all $a_k\;$ and, adding up the $n\;$ inequalities up, obtain

$\displaystyle\ln\left(\prod_{k=1}^{n}a_{k}^{a_k}\right)+\frac{1}{e}\sum_{k=1}^{n}e^{a_k}-\sum_{k=1}^{n}a_{k}^2\ge 0,\;$ from which $\displaystyle\ln\left(\prod_{k=1}^{n}a_{k}^{a_k}\right)+\frac{1}{e}\sum_{k=1}^{n}e^{a_k}\ge\frac{2}{n-1}\sum_{1\le i\lt j\le n}a_{i}a_{j}.$

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny