Difference between revisions of "Vornicu-Schur Inequality"

Revision as of 13:37, 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>

@@ Line 7: / Line 7: @@
 :<math>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</math>
-The standard form of Schur's is the case of this inequality where <math>x=a</math>, <math>y=b</math>, <math>z=c</math>, <math>k = 1</math>, and <math>f(m) = m^r</math>.<ref>Vornicu, Valentin;  ''Olimpiada de Matematica... de la provocare la experienta''; GIL Publishing House; Zalau, Romania.</ref>
+The standard form of Schur's is the case of this inequality where <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].
-==Notes==
+==References==
-<references/>
+*Vornicu, Valentin;  ''Olimpiada de Matematica... de la provocare la experienta''; GIL Publishing House; Zalau, Romania.</ref>
 [[Category:Theorems]]
 [[Category:Inequality]]

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Difference between revisions of "Vornicu-Schur Inequality"

Revision as of 13:37, 30 March 2008

Theorem

External Links

References