Difference between revisions of "User:Temperal/Inequalities"

Revision as of 13:42, 25 December 2007

Problem (me): Prove for $x,y,z\in\mathbb{R}^ + ,x + y + z = \frac {1}{2}$ that \[ \left(\sum_{sym}\frac {x}{y}\right)\left(\sqrt {2x^2 + 2y^2 + 2z^2}\right)\ge 1 \]

Solution (Altheman): By AM-GM, $\sum_{sym}\ge 6$ , and my RMS-AM $\sqrt {2x^2 + 2y^2 + 2z^2}\ge 1$ , thus the inequality is true.

Solution (me): By Jensen's Inequality, $\sum_{sym}\ge \frac{1}{\sum\limits_{sym}x\sqrt{y}}$ and by the Cauchy-Schwartz Inequality, $\frac{1}{\sum\limits_{sym}x\sqrt{y}}\ge\frac{1}{\sqrt {(2x^2 + 2y^2 + 2z^2)(2x+2y+2z)}=\frac{1}{\sqrt {2x^2 + 2y^2 + 2z^2}}

@@ Line 1: / Line 1: @@
-<cmath>\left(\sum_{sym}\frac {x}{y}\right)\left(\sqrt {(2x^2 + 2y^2 + 2z^2)(2x + 2y + 2z)}\right)\ge 1,x,y,z\in\mathbb{R}^ + ,x + y + z = \frac {1}{2}</cmath>
+'''Problem (me):''' Prove for <math>x,y,z\in\mathbb{R}^ + ,x + y + z = \frac {1}{2}</math> that
+\[
+\left(\sum_{sym}\frac {x}{y}\right)\left(\sqrt {2x^2 + 2y^2 + 2z^2}\right)\ge 1
+\]
-<cmath>\sum_{cyc}\frac{a\sqrt{a^2+3bc}}{\sqrt{b+c}}\ge a + b + c,a,b,c\in \mathbb{R}^+</cmath>
-<cmath>\left(\sum_{sym} x\sqrt {y}\right)\left(\sum_{sym}\frac {x}{\sqrt {y}}}\right)\ge 1,\{x,y,z|x,y,z\in\mathbb{R}^{ + },x + y + z = \frac {1}{2}\}</cmath>
+'''Solution (Altheman):''' By [[AM-GM]], <math>\sum_{sym}\ge 6</math>, and my [[Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality|RMS-AM]] <math>\sqrt {2x^2 + 2y^2 + 2z^2}\ge 1</math>, thus the inequality is true.
+'''Solution (me):''' By [[Jensen's Inequality]], <math>\sum_{sym}\ge \frac{1}{\sum\limits_{sym}x\sqrt{y}}</math> and by the [[Cauchy-Schwartz Inequality]], $\frac{1}{\sum\limits_{sym}x\sqrt{y}}\ge\frac{1}{\sqrt {(2x^2 + 2y^2 + 2z^2)(2x+2y+2z)}=\frac{1}{\sqrt {2x^2 + 2y^2 + 2z^2}}

Page

Toolbox

Search

Difference between revisions of "User:Temperal/Inequalities"

Revision as of 13:42, 25 December 2007