Ta co:$\frac{1}{\sqrt{x(y+z)}}$=$\frac{\sqrt{2} }{\sqrt{2x(y+z)}}$$\geq\frac{\sqrt{2} }{\frac{2x+y+z}{2}}=\frac{2\sqrt{2} }{2x+y+z}$$P\geq2\sqrt{2}(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z})$
$\geq2\sqrt{2}\frac{9}{2x+y+z+x+2y+z+x+y+2z}=\frac{18\sqrt{2}}{4(x+y+z)}=\frac{1}{4}$ $(x+y+z=18\sqrt{2})$
Dau"="$\Leftrightarrow x=y=z=6\sqrt{2}$