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})(xy+yz+zx)} $=\frac{(\sqrt{3}+1)xyz}{(xy+yz+zx)(\sqrt{x^{2}+y^{2}+z^{2}})} mà (xy+yz+zx)(\sqrt{x^{2}+y^{2}+z^{2}} \geq 3\sqrt[3]{x^{2}y^{2}z^{2}}\sqrt{3\sqrt[3]{x^{2}y^{2}z^{2}}}=3\sqrt{3}xyz \Rightarrow P\leq \frac{\sqrt{3}+1}{3\sqrt{3}}=\frac{3+\sqrt{3}}{9} dấu '=" \Leftrightarrow a=b=c$
C2) ta có $3(x^{2}+y^{2}+z^{2}
)\geq (x+y+z)^{2} \Rightarrow x+y+z \leq \sqrt{3(x^{2}+y^{2}+z^{2}
)}
\Rightarrow P\leq \frac{xyz(\sqrt{3(x^{2}+y^{2}+z^{2}}+\sqrt{x^{2}+y^{2}+z^{2}})}{(x^{2}+y^{2}+z^{2})
2(xy+yz+zx)}$
=$\frac{(\sqrt{3}+1)xyz}{
2(xy+yz+zx)\sqrt{x^{2}+y^{2}+z^{2}}}
mà (xy+yz+zx)\sqrt{x^{2}+y^{2}+z^{2}} \geq 3\sqrt[3]{x^{2}y^{2}z^{2}}\sqrt{3\sqrt[3]{x^{2}y^{2}z^{2}}}=3\sqrt{3}xyz
\Rightarrow P\leq \frac{\sqrt{3}+1}{
2.3\sqrt{3}}=\frac{3+\sqrt{3}}{
18}
dấu '=" \Leftrightarrow a=b=c$