\(\Leftrightarrow x+y=\sqrt{x+6}+\sqrt{y+6}\)
\(\Leftrightarrow\left(x+y\right)^2=\left(\sqrt{x+6}+\sqrt{y+6}\right)^2\le\left(1+1\right)\left(x+6+y+6\right)\) (BĐT cosi)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)-24\le0\)
Ta đặt \(x+y=m\)
\(m^2-2m-24\le0\)
\(\Leftrightarrow-4\le m\le6\)
\(\Rightarrow P_{max}=6\Leftrightarrow x=y=3\)