\({\log _{\sqrt 3 }}\frac{{x + y}}{{{x^2} + {y^2} + xy + 2}} = x\left( {x - 3} \right) + y\left( {y - 3} \right) + xy\,\,\,\,\,\,\left( 1 \right)\)

\( \Leftrightarrow {\log _{\sqrt 3 }}\left( {x + y} \right) - {\log _{\sqrt 3 }}\left( {{x^2} + {y^2} + xy + 2} \right) = {x^2} - 3x + {y^2} - 3y + xy\)

\( \Leftrightarrow {\log _{\sqrt 3 }}\left( {x + y} \right) + 3x + 3y = {\log _{\sqrt 3 }}\left( {{x^2} + {y^2} + xy + 2} \right) + {x^2} + {y^2} + xy\)

\(\begin{array}{l} \Leftrightarrow {\log _{\sqrt 3 }}\left( {x + y} \right) + 2 + 3x + 3y = {\log _{\sqrt 3 }}\left( {{x^2} + {y^2} + xy + 2} \right) + {x^2} + {y^2} + xy + 2\\ \Leftrightarrow {\log _{\sqrt 3 }}\left( {3x + 3y} \right) + 3x + 3y = {\log _{\sqrt 3 }}\left( {{x^2} + {y^2} + xy + 2} \right) + {x^2} + {y^2} + xy + 2\,\,\,\,\,\,\left( 2 \right) \end{array}\)

Đặt \(f\left( t \right) = {\log _{\sqrt 3 }}t + t,t > 0 \Rightarrow f'\left( t \right) = \frac{1}{{t\ln 3}} + 1 > 0,\forall t > 0 \Rightarrow f\left( t \right)\) đồng biến trên \(\left( {0; + \infty } \right)\)

\(\begin{array}{l} \left( 2 \right) \Leftrightarrow f\left( {3x + 3y} \right) = f\left( {{x^2} + {y^2} + xy + 2} \right) \Leftrightarrow 3x + 3 = {x^2} + {y^2} + xy + 2\\ \Leftrightarrow 4{x^2} + 4{y^2} + 4xy - 12x - 12y + 8 = 0\\ \Leftrightarrow {\left( {2x + y} \right)^2} - 6\left( {2x + y} \right) + 5 = - 3{\left( {y - 1} \right)^2} \le 0 \Leftrightarrow 1 \le 2x + y \le 5 \end{array}\)

Khi đó \(P = \frac{{3x + 2y + 1}}{{x + y + 6}} = 1 + \frac{{2x + y - 5}}{{x + y + 6}} \le 1\) vì \(\left\{ \begin{array}{l} 2x + y - 5 \le 0\\ x + y + 6 > 0 \end{array} \right.\)

Vậy P_max = 1 khi và chỉ khi \(\left\{ \begin{array}{l} 2x + y - 5 = 0\\ y - 1 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 2\\ y = 1 \end{array} \right.\)