Difference between revisions of "Vornicu-Schur Inequality"

(External Links)
 
(3 intermediate revisions by 3 users not shown)
Line 3: Line 3:
 
==Statement==
 
==Statement==
 
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
 
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
:<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>
+
<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>
 +
 
 +
[[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>.
 +
 
 +
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>
  
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>.
 
 
==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.
 
 
  
[[Category:Theorems]]
+
[[Category:Algebra]]
[[Category:Inequality]]
+
[[Category:Inequalities]]

Latest revision as of 16:58, 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.