Difference between revisions of "Schur's Inequality"

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'''Schur's Inequality''' is an [[inequality]] that holds for [[positive number]]s. It is named for Issai Schur.
 
'''Schur's Inequality''' is an [[inequality]] that holds for [[positive number]]s. It is named for Issai Schur.
  
 
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== Theorem ==
== THEOREM ==
 
 
Schur's inequality states that for all non-negative <math>a,b,c \in \mathbb{R}</math> and <math>r>0</math>:
 
Schur's inequality states that for all non-negative <math>a,b,c \in \mathbb{R}</math> and <math>r>0</math>:
  
<math>{a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b) \geq 0}</math>
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<cmath>a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b) \geq 0</cmath>
  
 
The four [[equality condition | equality cases]] occur when <math>a=b=c</math> or when two of <math>a,b,c</math> are equal and the third is <math>{0}</math>.
 
The four [[equality condition | equality cases]] occur when <math>a=b=c</math> or when two of <math>a,b,c</math> are equal and the third is <math>{0}</math>.
 
  
 
=== Common Cases ===
 
=== Common Cases ===
 
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The <math>r=1</math> case yields the well-known inequality:
The <math>r=1</math> case yields the well-known inequality:<math>a^3+b^3+c^3+3abc \geq a^2 b+a^2 c+b^2 a+b^2 c+c^2 a+c^2 b</math>
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<cmath>a^3+b^3+c^3+3abc \ge a^2 b+a^2 c+b^2 a+b^2 c+c^2 a+c^2 b</cmath>
  
 
When <math>r=2</math>, an equivalent form is:
 
When <math>r=2</math>, an equivalent form is:
<math>a^4+b^4+c^4+abc(a+b+c) \geq a^3 b+a^3 c+b^3 a+b^3 c+c^3 a+c^3 b</math>
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<cmath>a^4+b^4+c^4+abc(a+b+c) \ge a^3 b+a^3 c+b^3 a+b^3 c+c^3 a+c^3 b</cmath>
 
 
  
 
=== Proof ===
 
=== Proof ===
  
[[WLOG(Without loss of Generality)]], let <math>{a \geq b \geq c}</math>.  Note that <math>a^r(a-b)(a-c)+b^r(b-a)(b-c)</math> <math>= a^r(a-b)(a-c)-b^r(a-b)(b-c) = (a-b)(a^r(a-c)-b^r(b-c))</math>.  Clearly, <math>a^r \geq b^r \geq 0</math>, and <math>a-c \geq b-c \geq 0</math>. Thus, <math>(a-b)(a^r(a-c)-b^r(b-c)) \geq 0 \implies a^r(a-b)(a-c)+b^r(b-a)(b-c) \geq 0</math>.  However, <math>c^r(c-a)(c-b) \geq 0</math>, and thus the proof is complete.
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Without loss of Generality, let <math>{a\ge b\ge c}</math>.  Note that <math>a^r(a-b)(a-c)+b^r(b-a)(b-c)</math> <math>= a^r(a-b)(a-c)-b^r(a-b)(b-c) = (a-b)(a^r(a-c)-b^r(b-c))</math>.  Clearly, <math>a^r\ge b^r \ge 0</math>, and <math>a-c \geq b-c \geq 0</math>. Thus, <math>(a-b)(a^r(a-c)-b^r(b-c)) \geq 0 \implies a^r(a-b)(a-c)+b^r(b-a)(b-c) \geq 0</math>.  However, <math>c^r(c-a)(c-b) \geq 0</math>, and thus the proof is complete.
  
 
=== Generalized Form ===
 
=== Generalized Form ===
 
 
It has been shown by [[Valentin Vornicu]] that a more general form of Schur's Inequality exists.  Consider <math>a,b,c,x,y,z \in \mathbb{R}</math>, where <math>{a \geq b \geq 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 or monotonic.  Then,
 
It has been shown by [[Valentin Vornicu]] that a more general form of Schur's Inequality exists.  Consider <math>a,b,c,x,y,z \in \mathbb{R}</math>, where <math>{a \geq b \geq 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 or monotonic.  Then,
 
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<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 \geq 0.</cmath>
<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 \geq 0}</math>.
 
  
 
The standard form of Schur's is the case of this inequality where <math>x=a,\ y=b,\ z=c,\ k=1,\ f(m)=m^r</math>.
 
The standard form of Schur's is the case of this inequality where <math>x=a,\ y=b,\ z=c,\ k=1,\ f(m)=m^r</math>.
  
 
== References ==
 
== References ==
 
 
* Mildorf, Thomas; ''Olympiad Inequalities'';  January 20, 2006;  <http://artofproblemsolving.com/articles/files/MildorfInequalities.pdf>
 
* Mildorf, Thomas; ''Olympiad Inequalities'';  January 20, 2006;  <http://artofproblemsolving.com/articles/files/MildorfInequalities.pdf>
  

Latest revision as of 18:19, 17 June 2021

Schur's Inequality is an inequality that holds for positive numbers. It is named for Issai Schur.

Theorem

Schur's inequality states that for all non-negative $a,b,c \in \mathbb{R}$ and $r>0$:

\[a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b) \geq 0\]

The four equality cases occur when $a=b=c$ or when two of $a,b,c$ are equal and the third is ${0}$.

Common Cases

The $r=1$ case yields the well-known inequality: \[a^3+b^3+c^3+3abc \ge a^2 b+a^2 c+b^2 a+b^2 c+c^2 a+c^2 b\]

When $r=2$, an equivalent form is: \[a^4+b^4+c^4+abc(a+b+c) \ge a^3 b+a^3 c+b^3 a+b^3 c+c^3 a+c^3 b\]

Proof

Without loss of Generality, let ${a\ge b\ge c}$. Note that $a^r(a-b)(a-c)+b^r(b-a)(b-c)$ $= a^r(a-b)(a-c)-b^r(a-b)(b-c) = (a-b)(a^r(a-c)-b^r(b-c))$. Clearly, $a^r\ge b^r \ge 0$, and $a-c \geq b-c \geq 0$. Thus, $(a-b)(a^r(a-c)-b^r(b-c)) \geq 0 \implies a^r(a-b)(a-c)+b^r(b-a)(b-c) \geq 0$. However, $c^r(c-a)(c-b) \geq 0$, and thus the proof is complete.

Generalized Form

It has been shown by Valentin Vornicu that a more general form of Schur's Inequality exists. Consider $a,b,c,x,y,z \in \mathbb{R}$, where ${a \geq b \geq c}$, 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, \[f(x)(a-b)^k(a-c)^k+f(y)(b-a)^k(b-c)^k+f(z)(c-a)^k(c-b)^k \geq 0.\]

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

References

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

See Also

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