Ta có \((x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\Rightarrow (x+y+z)^2=3+2(xy+yz+zx)\)
lại có \(x^3+y^3+z^3=(x+y+z)\left [ x^2+y^2+z^2-(xy+yz+zx) \right ]+3xyz\) \(=(x+y+z)\left [ 3-(xy+yz+zx) \right ]+3xyz\) nên \(\frac{x^3+y^3+z^3}{9xyz}=\frac{1}{3}+\frac{1}{9}\left ( \frac{1}{yz}+\frac{1}{zx}+\frac{1}{xy} \right )\left [ 3-(xy+yz+zx) \right ]\) Áp dụng BĐT Cauchy ta có \(\left\{\begin{matrix} xy+yz+zx\geq 3\sqrt[3]{x^2.y^2.z^2}\\ \frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\geq 3\sqrt[3]{\frac{1}{x^2.y^2.z^2}} \end{matrix}\right.\Rightarrow \frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\geq \frac{9}{xy+yz+zx}\) Suy ra \(\frac{x^3+y^3+z^3}{9xyz}\geq \frac{1}{3}+\left ( \frac{1}{xy+yz+zx} \right )\left [ 3-(xy+yz+zx) \right ]\) Từ đó ta có \(P\leq 3+2(xy+yz+zx)-\frac{1}{3}-\left (\frac{1}{xy+yz+zx} \right )\left [ 3-(xy+yz+zx) \right ]\)\(+\frac{3}{xy+yz+zx}\) \(=\frac{11}{3}+2(xy+yz+zx)\) do \(0 Từ đó suy ra GTLN của P là \(\frac{29}{3}\) đạt khi \(\left\{\begin{matrix} x^2+y^2+z^2=3\\ xy=yz=xz\\ xy+yz+zx=3 \end{matrix}\right.\Leftrightarrow x=y=z=1\)