\(\frac{4}{3}\ge x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\Rightarrow x+y+z\le2\)
\(\left\{{}\begin{matrix}x;y;z\ge0\\x^2+y^2+z^2=\frac{4}{3}\end{matrix}\right.\) \(\Rightarrow0\le x;y;z\le\frac{2\sqrt{3}}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}x\left(x-\frac{2\sqrt{3}}{3}\right)\le0\\y\left(y-\frac{2\sqrt{3}}{3}\right)\le0\\z\left(z-\frac{2\sqrt{3}}{3}\right)\le0\end{matrix}\right.\) \(\Rightarrow x+y+z\ge\frac{x^2+y^2+z^2}{\frac{2\sqrt{3}}{3}}=\frac{2\sqrt{3}}{3}\)
Đặt \(x+y+z=t\Rightarrow\frac{2\sqrt{3}}{3}\le t\le2\)
\(P=4t-\frac{3}{t}=\frac{8t^2-13t-6}{2t}+\frac{13}{2}=\frac{\left(t-2\right)\left(8t+3\right)}{2t}+\frac{13}{2}\le\frac{13}{2}\)
\(P_{max}=\frac{13}{2}\) khi \(t=2\) hay \(x=y=z=\frac{2}{3}\)
\(t\ge\frac{2\sqrt{3}}{3}\Rightarrow P=4t-\frac{3}{t}\ge4.\frac{2\sqrt{3}}{3}-\frac{3}{\frac{2\sqrt{3}}{3}}=\frac{7\sqrt{3}}{6}\)
\(P_{min}=\frac{7\sqrt{3}}{6}\) khi \(t=\frac{2\sqrt{3}}{3}\) hay \(\left(x;y;z\right)=\left(0;0;\frac{2\sqrt{3}}{3}\right)\) và hoán vị
