Ta có: $\mathbb P=\frac{1}{\frac{1}{x}+\frac{1}{y}+\frac{1}{y}}+\frac{1}{\frac{1}{3z}+\frac{1}{3z}+\frac{1}{3y}}+\frac{1}{\frac{1}{6x}+\frac{1}{6x}+\frac{1}{6z}}$Theo bất đẳng thức AM - GM, ta có: $(x+y+y)(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}) \ge 9\Leftrightarrow \frac{x+y+y}{9} \ge \frac{1}{\frac{1}{x}+\frac{1}{y}+\frac{1}{y}}$ $(y+z+z)(\frac{1}{3z}+\frac{1}{3z}+\frac{1}{3y}) \ge 3\Leftrightarrow \frac {y+z+z}{3} \ge \frac{1}{\frac{1}{3z}+\frac{1}{3z}+\frac{1}{3y}}$ $(z+x+x)(\frac{1}{6x}+\frac{1}{6x}+\frac{1}{6z}) \ge \frac{3}{2}\Leftrightarrow \frac {2(z+x+x)}{3} \ge \frac{1}{\frac{1}{6x}+\frac{1}{6x}+\frac{1}{6z}}$ Cộng theo vế các bất đẳng thức trên ta được: $\mathbb P \le (\frac{1}{9}+\frac{4}{3})x+(\frac{2}{9}+\frac{1}{3})y+(\frac{2}{3}+\frac{2}{3})z=\frac{1}{9} (13x+5y+12z)=1$Vậy: $\color{red}{\boxed{\max P=1}}$ tại $\color{red}{x=y=z=\frac{3}{10}.}$
Ta có: $\mathbb P=\frac{1}{\frac{1}{x}+\frac{1}{y}+\frac{1}{y}}+\frac{1}{\frac{1}{3z}+\frac{1}{3z}+\frac{1}{3y}}+\frac{1}{\frac{1}{6x}+\frac{1}{6x}+\frac{1}{6z}}$Theo bất đẳng thức AM - GM, ta có: $(x+y+y)(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}) \ge 9\Leftrightarrow \frac{x+y+y}{9} \ge \frac{1}{\frac{1}{x}+\frac{1}{y}+\frac{1}{y}}$ $(y+z+z)(\frac{1}{3z}+\frac{1}{3z}+\frac{1}{3y}) \ge 3\Leftrightarrow \frac {y+z+z}{3} \ge \frac{1}{3z}+\frac{1}{3z}+\frac{1}{3y}$ $(z+x+x)(\frac{1}{6x}+\frac{1}{6x}+\frac{1}{6z}) \ge \frac{3}{2}\Leftrightarrow \frac {2(z+x+x)}{3} \ge \frac{1}{6x}+\frac{1}{6x}+\frac{1}{6z}$ Cộng theo vế các bất đẳng thức trên ta được: $\mathbb P \le (\frac{1}{9}+\frac{4}{3})x+(\frac{2}{9}+\frac{1}{3})y+(\frac{2}{3}+\frac{2}{3})z=\frac{1}{9}(13x+5y+12z)=1$Vậy: $\color{red}{\boxed{\max P=1}}$ tại $\color{red}{x=y=z=\frac{3}{10}.}$
Ta có: $\mathbb P=\frac{1}{\frac{1}{x}+\frac{1}{y}+\frac{1}{y}}+\frac{1}{\frac{1}{3z}+\frac{1}{3z}+\frac{1}{3y}}+\frac{1}{\frac{1}{6x}+\frac{1}{6x}+\frac{1}{6z}}$Theo bất đẳng thức AM - GM, ta có: $(x+y+y)(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}) \ge 9\Leftrightarrow \frac{x+y+y}{9} \ge \frac{1}{\frac{1}{x}+\frac{1}{y}+\frac{1}{y}}$ $(y+z+z)(\frac{1}{3z}+\frac{1}{3z}+\frac{1}{3y}) \ge 3\Leftrightarrow \frac {y+z+z}{3} \ge \frac{1}{
\frac{1}{3z}+\frac{1}{3z}+\frac{1}{3y
}}$ $(z+x+x)(\frac{1}{6x}+\frac{1}{6x}+\frac{1}{6z}) \ge \frac{3}{2}\Leftrightarrow \frac {2(z+x+x)}{3} \ge \frac{1}{
\frac{1}{6x}+\frac{1}{6x}+\frac{1}{6z}
}$
Cộng theo vế các bất đẳng thức trên ta được: $\mathbb P \le (\frac{1}{9}+\frac{4}{3})x+(\frac{2}{9}+\frac{1}{3})y+(\frac{2}{3}+\frac{2}{3})z=\frac{1}{9}
(13x+5y+12z)=1$Vậy: $\color{red}{\boxed{\max P=1}}$ tại $\color{red}{x=y=z=\frac{3}{10}.}$