Difference between revisions of "Vornicu-Schur Inequality"

m
(External Links)
Line 6: Line 6:
  
 
Schur's Inequality follows from Vornicu-Schur by setting <math>x=a</math>, <math>y=b</math>, <math>z=c</math>, <math>k = 1</math>, and <math>f(m) = m^r</math>.
 
Schur's Inequality follows from Vornicu-Schur by setting <math>x=a</math>, <math>y=b</math>, <math>z=c</math>, <math>k = 1</math>, and <math>f(m) = m^r</math>.
==External Links==
 
*A full statement, as well as some applications can be found in [http://www.mathlinks.ro/portal.php?t=162684 this article].
 
 
 
==References==
 
==References==
 
*Vornicu, Valentin;  ''Olimpiada de Matematica... de la provocare la experienta''; GIL Publishing House; Zalau, Romania.</ref>
 
*Vornicu, Valentin;  ''Olimpiada de Matematica... de la provocare la experienta''; GIL Publishing House; Zalau, Romania.</ref>

Revision as of 15:53, 4 September 2010

The Vornicu-Schur Inequality is a generalization of Schur's Inequality discovered by the Romanian mathematician Valentin Vornicu.

Statement

Consider real numbers $a,b,c,x,y,z$ such that $a \ge b \ge c$ and either $x \geq y \geq z$ or $z \geq y \geq x$. Let $k \in \mathbb{Z}_{> 0}$ be a positive integer and let $f:\mathbb{R} \rightarrow \mathbb{R}_{\geq 0}$ be a function from the reals to the nonnegative reals that is either convex or monotonic. Then

$f(x)(a-b)^k(a-c)^k+f(y)(b-a)^k(b-c)^k+f(z)(c-a)^k(c-b)^k \ge 0.$

Schur's Inequality follows from Vornicu-Schur by setting $x=a$, $y=b$, $z=c$, $k = 1$, and $f(m) = m^r$.

References

  • Vornicu, Valentin; Olimpiada de Matematica... de la provocare la experienta; GIL Publishing House; Zalau, Romania.</ref>