Difference between revisions of "Vornicu-Schur Inequality"

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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>.
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The most widely used form of Vornicu-Schur is
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Let <math>a \geq b \geq c</math> be real numbers, and let <math>x,y,z \geq 0</math>. Then if <math>x+z\geq y</math> then the following inequality is true
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<cmath> x(a-b)(a-c) + y(b-c)(b-a) + z (c-a)(c-b) \geq 0 . </cmath>
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==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 15:54, 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 $f(m) = m^r$.

The most widely used form of Vornicu-Schur is

Let $a \geq b \geq c$ be real numbers, and let $x,y,z \geq 0$. Then if $x+z\geq y$ then the following inequality is true \[x(a-b)(a-c) + y(b-c)(b-a) + z (c-a)(c-b) \geq 0 .\]

References

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