Xét các số thực dươg x, y  thỏa mãn ({log _{sqrt 3 }}frac{{x + y}}{{{x^2} + {y^2} + xy + 2}} = xleft( {x – 3} right) + yleft(

({log _{sqrt 3 }}frac{{x + y}}{{{x^2} + {y^2} + xy + 2}} = xleft( {x – 3} right) + yleft( {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 (fleft( t right) = {log _{sqrt 3 }}t + t,t > 0 Rightarrow f’left( t right) = frac{1}{{tln 3}} + 1 > 0,forall t > 0 Rightarrow fleft( t right)) đồng biến trên (left( {0; + infty } right))

(begin{array}{l}
left( 2 right) Leftrightarrow fleft( {3x + 3y} right) = fleft( {{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} – 6left( {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.)