\(\left(x+3y\right)^2\le\left(1+3^2\right)\left(x^2+y^2\right)=10\left(x^2+y^2\right)\)
\(\Rightarrow5\left(x^2+y^2\right)\ge\frac{1}{2}\left(x+3y\right)^2\)
\(\Rightarrow\frac{1}{2}\left(x+3y\right)^2-5\left(x+3y\right)+8\le0\)
\(\Leftrightarrow\left(x+3y\right)^2-10\left(x+3y\right)+16\le0\)
\(\Rightarrow2\le x+3y\le8\)
\(\Rightarrow3\le x+3y+1\le9\)