\(x^2+y^2=1\Leftrightarrow2xy=\left(x+y\right)^2-1\Rightarrow xy=\frac{1}{2}\left(x+y\right)^2-\frac{1}{2}\)
Mặt khác \(1=x^2+y^2\ge\frac{1}{2}\left(x+y\right)^2\Rightarrow-\sqrt{2}\le x+y\le\sqrt{2}\)
\(P=9+xy-3\left(x+y\right)=9+\frac{1}{2}\left(x+y\right)^2-\frac{1}{2}-3\left(x+y\right)\)
\(2P=\left(x+y\right)^2-6\left(x+y\right)+17\)
\(2P=\left(x+y\right)^2-2-6\left(x+y\right)+6\sqrt{2}+19-6\sqrt{2}\)
\(2P=\left(x+y+\sqrt{2}\right)\left(x+y-\sqrt{2}\right)-6\left(x+y-\sqrt{2}\right)+19-6\sqrt{2}\)
\(2P=\left(x+y-\sqrt{2}\right)\left(x+y-6+\sqrt{2}\right)+19-6\sqrt{2}\ge19-6\sqrt{2}\)
\(\Rightarrow P\ge\frac{19-6\sqrt{2}}{2}\)
\(P_{min}=\frac{19-6\sqrt{2}}{2}\) khi \(x=y=\frac{\sqrt{2}}{2}\)
