User:Temperal/Inequalities

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

Solution (Altheman): By AM-GM, , and my RMS-AM , thus the inequality is true.

Solution (me): By Jensen's Inequality, 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}}