Ta có : \(\hept{\begin{cases}0\le a\le2\\0\le b\le2\\0\le c\le2\end{cases}}\)\(\Rightarrow\hept{\begin{cases}a\left(2-a\right)\ge0\\b\left(2-b\right)\ge0\\c\left(2-c\right)\ge0\end{cases}}\)
\(\Rightarrow-a^2+2a-b^2+2b-c^2+2c\ge0\)
\(\Leftrightarrow a^2+b^2+c^2\le2\left(a+b+c\right)=2.3=6\)
Vậy Max P = 6