Difference between revisions of "Vornicu-Schur Inequality"

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==Theorem==
 
==Theorem==
In [[2007]], [[Romanian]] mathematician [[Valentin Vornicu]] showed that a generalized form of Schur's inequality exists:  
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In 2007, Romanian mathematician [[Valentin Vornicu]] showed that a generalized form of Schur's inequality exists:  
  
 
Consider <math>a,b,c,x,y,z \in \mathbb{R}</math>, where <math>a \ge b \ge c</math>, and either <math>x \geq y \geq z</math> or <math>>z \geq y \geq x</math>.  Let <math>k \in \mathbb{Z}^{+}</math>, and let <math>f:\mathbb{R} \rightarrow \mathbb{R}_{0}^{+}</math> be either [[convex function|convex]] or [[monotonic]].  Then,
 
Consider <math>a,b,c,x,y,z \in \mathbb{R}</math>, where <math>a \ge b \ge c</math>, and either <math>x \geq y \geq z</math> or <math>>z \geq y \geq x</math>.  Let <math>k \in \mathbb{Z}^{+}</math>, and let <math>f:\mathbb{R} \rightarrow \mathbb{R}_{0}^{+}</math> be either [[convex function|convex]] or [[monotonic]].  Then,

Revision as of 13:38, 30 March 2008

The Vornicu-Schur' refers to a generalized version of Schur's Inequality.

Theorem

In 2007, Romanian mathematician Valentin Vornicu showed that a generalized form of Schur's inequality exists:

Consider $a,b,c,x,y,z \in \mathbb{R}$, where $a \ge b \ge c$, and either $x \geq y \geq z$ or $>z \geq y \geq x$. Let $k \in \mathbb{Z}^{+}$, and let $f:\mathbb{R} \rightarrow \mathbb{R}_{0}^{+}$ be either convex or monotonic. Then,

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

The standard form of Schur's is the case of this inequality where $x=a$, $y=b$, $z=c$, $k = 1$, and $f(m) = m^r$.

External Links

  • A full statement, as well as some applications can be found in this article.

References

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