B1
[tex]\frac{1}{\sum x^2}+\frac{1}{xyz}=\frac{1}{\sum x^2}+\frac{x+y+z}{xyz}=\frac{1}{\sum x^2}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{1}{\sum x^2}+\sum\frac{1}{3xy}+\sum\frac{2}{3xy}\geq \frac{4^2}{(\sum x)^2+\sum xy}+2.\frac{3^2}{3\sum xy}\geq \frac{16}{1+\frac{(\sum x)^2}{3}}+\frac{18}{(\sum x)^2}=12+18=30[/tex]
B2
[tex]GT\Leftrightarrow \sum \frac{1}{xy}=1\\\sum \frac{1+\sqrt{1+x^2}}{x}=\sum \frac{1}{x}+\sum \sqrt{\frac{1}{x^2}+1}=\sum \frac{1}{x}+\sum \sqrt{\frac{1}{x^2}+\sum \frac{1}{xy}}=\sum \frac{1}{x}+\sum \sqrt{(\frac{1}{x}+\frac{1}{y})(\frac{1}{x}+\frac{1}{z})}\leq \sum \frac{1}{x}+\sum \frac{\frac{1}{x}+\frac{1}{z}+\frac{1}{x}+\frac{1}{y}}{2}=\sum \frac{3}{x}=\frac{3\sum xy}{xyz}\leq \frac{(\sum x)^2}{xyz}=\frac{(xyz)^2}{xyz}=xyz[/tex]