Thay \(c=2-\left(a+b\right)\Leftrightarrow P=2ab+c\left(a+b\right)=2ab+\left(a+b\right)\left[2-\left(a+b\right)\right]\)
\(=2ab+2\left(a+b\right)-a^2-b^2-2ab=2\left(a+b\right)-a^2-b^2=2-\left(a-1\right)^2-\left(b-1\right)^2\)
Mà \(\hept{\begin{cases}\left(a-1\right)^2\\\left(b-1\right)^2\end{cases}\ge0\forall a,b\inℝ\Rightarrow P=2-\left(a-1\right)^2-\left(b-1\right)^2\le2}\)
Dấu ''='' xảy ra \(\Leftrightarrow\) \(a=b=1\rightarrow c=0\)