đk: \(x,y\ge-6\Rightarrow x+y\ge0\)
Theo bài ra, ta có:
\(\left(x+y\right)^2=\left(\sqrt{x+6}+\sqrt{y+6}\right)^2\)
\(=x+y+12+2\sqrt{\left(x+6\right)\left(y+6\right)}\le x+y+12+x+6+y+6\)
Hay \(\left(x+y\right)^2\le2\left(x+y\right)+24\)
\(\Leftrightarrow\left(x+y+4\right)\left(x+y-6\right)\le0\)
\(\Leftrightarrow x+y-6\le0\Leftrightarrow x+y\le6\)
Dấu '=' xảy ra<=> x=y=3
=> GTNN của P là 6 <=> x=y=3
Đặt \(a=\sqrt{x+6};b=\sqrt{y+6}\Rightarrow a;b\ge0,a+b=a^2+b^2-12\)
và \(P=a^2+b^2-12=a+b\)
Ta có: \(a+b=\left(a+b\right)^2-2ab-12\Rightarrow a+b\le\left(a+b\right)^2-12\left(a;b\ge0\right)\)
Hay \(\left(a+b\right)^2-\left(a+b\right)-12\ge0\)
\(\Leftrightarrow\left(a+b+3\right)\left(a+b-4\right)\ge0\)
\(\Leftrightarrow a+b\ge4\Rightarrow P\ge4\)
Dấu '=' xảy ra <=> \(\hept{\begin{cases}a=4\\b=0\end{cases}}\) hoặc \(\hept{\begin{cases}a=0\\b=4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=10\\y=-6\end{cases}}\) hoặc \(\hept{\begin{cases}x=-6\\y=10\end{cases}}\)
