A Not Symmetric Inequality

Problem

Prove the inequality

$\displaystyle \sum_{i=1}^{n}\frac{x_{i}^{3}}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}\ge\frac{1}{3}\sum_{i=1}^{n}x_{i},$

where $x_{1},x_{2},\ldots,x_{n}$ are all positive and $x_{n+1}=x_{1}.$

Solution

The summands on the left lack symmetry which seems to be desirable if not easy to repair. How would the repaired left-hand side look like?

$\displaystyle \sum_{i=1}^{n}\frac{x_{i}^{3}+x_{i+1}^{3}}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}.$

But, since $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$ and $\displaystyle\frac{a^{2}-ab+b^{2}}{a^{2}+ab+b^{2}}\ge\frac{1}{3},$ the modified left-hand side would allow an estimate

$\displaystyle \sum_{i=1}^{n}\frac{x_{i}^{3}+x_{i+1}^{3}}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}\ge \frac{2}{3}\sum_{i=1}^{n}x_{i}.$

Now, the perhaps unexpected part of this is that there is an identity:

$\displaystyle \sum_{i=1}^{n}\frac{x_{i}^{3}}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}=\sum_{i=1}^{n}\frac{x_{i+1}^{3}}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}},$

such that the modified left-hand side was equal to exactly twice the original one, and this solves the problem. But why we have that identity? This is due to $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$ such that

$\displaystyle \sum_{i=1}^{n}\frac{x_{i}^{3}-x_{i+1}^{3}}{x_{i}^{2}+x_{i}x_{i+1}+x_{i+1}^{2}}=\sum_{i=1}^{n}(x_{i}-x_{i+1})=0.$

References

  1. S. Savchev, T. Andreescu, Mathematical Miniatures, MAA, 2003, pp. 73-74

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

Copyright © 1996-2017 Alexander Bogomolny

 62596656

Search by google: