Difference between revisions of "User:Temperal/Inequalities"

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

Latest revision as of 12:01, 11 June 2008

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-Schwarz Inequality, $\frac{1}{\sum\limits_{sym}x\sqrt{y}}\ge\frac{1}{\sqrt{(2x^2 + 2y^2 + 2z^2)(2x+2y+2z)}}=\frac{1}{\sqrt{2x^2 + 2y^2 + 2z^2}}$