\(\left|\left(x-3\right)+2\left(y-1\right)\right|\le\sqrt{\left(1+4\right)\left[\left(x-3\right)^2+\left(y-1\right)^2\right]}=5\)
\(\Rightarrow-5\le x+2y-5\le5\Rightarrow0\le x+2y\le10\)
\(P=\frac{x^2+4y^2+4xy+x+2y+9-\left(x^2-6x+9\right)-\left(y^2-2y+1\right)}{x+2y+1}\)
\(P=\frac{\left(x+2y\right)^2+\left(x+2y\right)+9-\left(x-3\right)^2-\left(y-1\right)^2}{x+2y+1}=\frac{\left(x+2y\right)^2+\left(x+2y\right)+4}{x+2y+1}\)
Đặt \(x+2y=t\ge0\)
\(P=\frac{t^2+t+4}{t+1}=t+\frac{4}{t+1}=t+1+\frac{4}{t+1}-1\)
\(P\ge2\sqrt{\frac{4\left(t+1\right)}{t+1}}-1=3\)
