Difference between revisions of "User:Temperal/Inequalities"
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\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> | </cmath> | ||
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| + | '''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. | ||
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| + | '''Solution (me):''' By [[Jensen's Inequality]], <math>\sum_{sym}\ge \frac{1}{\sum\limits_{sym}x\sqrt{y}}</math> and by the [[Cauchy-Schwarz Inequality]], <math>\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}}</math> | ||
Latest revision as of 12:01, 11 June 2008
Problem (me): Prove for 
that
![\[\left(\sum_{sym}\frac {x}{y}\right)\left(\sqrt {2x^2 + 2y^2 + 2z^2}\right)\ge 1\]](http://latex.artofproblemsolving.com/3/b/a/3ba61b4841f12f8037f32be749db2374e9e638c6.png)
Solution (Altheman): By AM-GM, 
, and my RMS-AM 
, thus the inequality is true.
Solution (me): By Jensen's Inequality, 
and by the Cauchy-Schwarz Inequality, 
