C2) ta có $3(x^{2}+y^{2}+z^{2})\geq (x+y+z)^{2} \Rightarrow x+y+z \leq \sqrt{3(x^{2}+y^{2}+z^{2})}$
$\Rightarrow P\leq \frac{xyz(\sqrt{3(x^{2}+y^{2}+z^{2}}+\sqrt{x^{2}+y^{2}+z^{2}})}{(x^{2}+y^{2}+z^{2})2(xy+yz+zx)}$
=$\frac{(\sqrt{3}+1)xyz}{2(xy+yz+zx)\sqrt{x^{2}+y^{2}+z^{2}}}$
mà $(xy+yz+zx)\sqrt{x^{2}+y^{2}+z^{2}} \geq 3\sqrt[3]{x^{2}y^{2}z^{2}}\sqrt{3\sqrt[3]{x^{2}y^{2}z^{2}}}=3\sqrt{3}xyz$
$\Rightarrow P\leq \frac{\sqrt{3}+1}{2.3\sqrt{3}}=\frac{3+\sqrt{3}}{18}$
dấu '=" $\Leftrightarrow a=b=c$