$9 \ge \bigg[(x+4y+\frac {17}5)-\frac {12}5 \bigg]^2+\bigg[(3x+7y-\frac 45)+\frac 95 \bigg]^2 \\ =\big(x+4y+\tfrac{17}5 \big)^2-\tfrac{24}{5}\bigl(x+4y+\tfrac {17}5 \big)+\tfrac{144}{25}+\big(3x+7y-\tfrac 45 \big)^2+\tfrac{18}{5}\big(3x+7y-\tfrac 45)+\tfrac{81}{25} \\ \ge \frac{18}{5}\left(3x+7y-\frac 45\right)-\frac{24}{5}\left(x+4y+\frac{17}5\right)+9\Leftrightarrow x+y \le \frac {16}5 \\ \mathbf{Dấu\;bằng\;xảy\;ra\;khi\;} x=\frac{27}5,y=-\frac{11}5 \\ \mathbf{Bằng\;cách\;tương\;tự\;cm\;đc\;} x+y \ge -\frac{14}5. \mathbf{Dấu\;bằng\;xảy\;ra\;khi\;} x=-\frac{21}5,y=\frac 75$