Difference between revisions of "Vornicu-Schur Inequality"

Latest revision as of 15:58, 29 December 2021

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 in the case $f(x) = x$ , $k = 1$ , when we have for real numbers $a \geq b \geq c$ and nonnegative real numbers $x, y, z$ that if $x + z \geq y$ then $x(a-b)(a-c) + y(b-c)(b-a) + z (c-a)(c-b) \geq 0 .$

References

Vornicu, Valentin; Olimpiada de Matematica... de la provocare la experienta; GIL Publishing House; Zalau, Romania.

@@ Line 1: / Line 1: @@
-The '''Vornicu-Schur'''' refers to a generalized version of [[Schur's Inequality]].
+The '''Vornicu-Schur Inequality''' is a generalization of [[Schur's Inequality]] discovered by the Romanian mathematician [[Valentin Vornicu]].
-==Theorem==
+==Statement==
-In [[2007]], [[Romanian]] mathematician [[Valentin Vornicu]] showed that a generalized form of Schur's inequality exists:
+Consider [[real number]]s <math>a,b,c,x,y,z</math> such that <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}_{> 0}</math> be a [[positive integer]] and let <math>f:\mathbb{R} \rightarrow \mathbb{R}_{\geq 0}</math> be a [[function]] from the reals to the nonnegative reals that is either [[convex function|convex]] or [[monotonic]].  Then
+<cmath>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.</cmath>
-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,
+[[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>.
-:<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 most widely used form of Vornicu-Schur is in the case <math>f(x) = x</math>, <math>k = 1</math>, when we have for real numbers <math>a \geq b \geq c</math> and nonnegative real numbers <math>x, y, z</math> that if <math>x + z \geq y</math> then
+<cmath> x(a-b)(a-c) + y(b-c)(b-a) + z (c-a)(c-b) \geq 0 . </cmath>
-==External Links==
+==References==
-*A full statement, as well as some applications can be found in [http://www.mathlinks.ro/portal.php?t=162684 this article].
+*Vornicu, Valentin;  ''Olimpiada de Matematica... de la provocare la experienta''; GIL Publishing House; Zalau, Romania.
-==Notes==
+[[Category:Algebra]]
-<references/>
+[[Category:Inequalities]]
-[[Category:Theorems]]
-[[Category:Inequality]]

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Difference between revisions of "Vornicu-Schur Inequality"

Latest revision as of 15:58, 29 December 2021

Statement

References