Cute Exercise by Dorin Marghidanu

Problem 1

Cute inequality by Dorin Marghidanu

Problem 2

Cute inequality by Dorin Marghidanu, problem 2

Solution 1 to Problem 2

Recollect that $\displaystyle \sum_{k=1}^n(2k-1)=n^2.$ We shall use the AM-GM inequality:

$\displaystyle\begin{align}\sum_{k=1}^n\frac{2k-1}{\sqrt[2k-1]{a_k}}&= \frac{1}{a_1}+\frac{1}{\sqrt[3]{a_2}}+\frac{1}{\sqrt[3]{a_2}}+\frac{1}{\sqrt[3]{a_2}}+\ldots+\underbrace{\frac{1}{\sqrt[2n-1]{a_n}}+\ldots+\frac{1}{\sqrt[2n-1]{a_n}}}_{2n-1\text{ times}}\\ &\ge\frac{n^2}{\sqrt[n^2]{a_1a_2\ldots a_n}}. \end{align}$

Equality is attained when $\sqrt{a_1}=\sqrt[3]{a_2}=\ldots=\sqrt[2n-1]{a_n}.$

Solution 2 to Problem 2

This time, in addition to the AM-GM inequality we'll use math induction.

For $n=1$ the inequality obviously holds (as an equality in fact). Assume it holds for some $n$ and deduce from that its validity for $n+1.$

$\displaystyle\begin{align}\sum_{k=1}^{n+1}\frac{2k-1}{\sqrt[2k-1]{a_k}}&=\sum_{k=1}^{n}\frac{2k-1}{\sqrt[2k-1]{a_k}}+\frac{2n+1}{\sqrt[2n+1]{a_{n+1}}}\\ &\ge\frac{n^2}{\sqrt[n^2]{a_1a_2\ldots a_n}}+\frac{2n+1}{\sqrt[2n+1]{a_{n+1}}}\\ &=\underbrace{\underbrace{\frac{1}{\sqrt[n^2]{a_1\ldots a_n}}+\ldots+\frac{1}{\sqrt[n^2]{a_1\ldots a_n}}}_{n^2\text{ times}}+\underbrace{\frac{1}{\sqrt[2n+1]{a_{n+1}}}+\ldots+\frac{1}{\sqrt[2n+1]{a_{n+1}}}}_{2n+1\text{ times}}}_{(n+1)^2\text{ terms in all}}\\ &\ge\frac{(n+1)^2}{\sqrt[(n+1)^2]{a_1a_2\ldots a_na_{n+1}}}. \end{align}$

Acknowledgment

Dorin Marghidanu has kindly communicated to me his problem (Problem 1) and its solution. I modified the problem slightly (Problem 2) and followed in Dorin's footsteps in proving the latter.

 

[an error occurred while processing this directive]

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

Copyright © 1996-2018 Alexander Bogomolny
[an error occurred while processing this directive]
[an error occurred while processing this directive]